Unit 01A - New Age International

UNIT – I

ACOUSTICS

AND

STRUCTURE

OF

SOLIDS A. ACOUSTICS 1.1 INTRODUCTION Acoustics refers to the branch of science which deals with audible sound. It is well known that sound is a form of energy. The propagation of sound in a medium, from one place to another, takes place in the form of wave motion. During the propagation of sound, the particles of the medium execute simple harmonic vibrations. The condition for propagation of wave motion in a medium is that the medium should possess inertia, elasticity and be capable of resisting the motion. It means that sound cannot travel in vacuum. The way particles of the medium vibrate with respect to the direction of propagation of sound results in either longitudinal or transverse wave motion. You are aware that longitudinal wave motion is one is which particles of a medium vibrate parallel to the direction of propagation of energy. Transverse wave motion is one in which particles of a medium vibrate in a direction perpendicular to the direction of propagation of enegy. More about these wave motions have been discussed in earlier classes of education.

1.2 CLASSIFICATION OF SOUND Basically sound may be classified as ‘music’ and ‘noise’. Music or musical sound is produced by wave motions which have certain regularity in them. Music produces pleasant sensation for humankind. Noise is an irregular sound and is not desired. Noise produces ill effects on human beings.

1.3 CHARACTERISTICS OF MUSICAL SOUND A musical sound travelling in a medium will produce simple harmonic motion of particles with certain regularity. These simple harmonic motions have a definite period, frequency and amplitude. There will not be any marked deviations in them. Following are the characteristics of musical sound.

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A Textbook of Engineering Physics

1. Pitch or frequency 2. Loudness 3. Quality or Timbre

1.3.1 Pitch The sensation produced by a sound in the brain of a person or listener is called pitch. Pitch differentiates between a shrill sound (that will be piercing with high intensity) and a grave sound (that is slow and impressive). Mosquitoes produce shrill sounds of high pitch whereas lions produce sounds of low pitch. Even though pitch depends upon the frequency of a sound it should be understood that pitch and frequency are different. As the frequency of a musical sound increases, the pitch of sound also increases. Similarly, lower is the frequency of a musical sound lower is the pitch. Pitch and frequency are two different quantities. Pitch is a physiological quantity because it is only a sensation a human being or a listener feels. Hence it cannot be measured by any instrument. But frequency is a physical quantity, namely, the number of vibrations or oscillations made in one second by vibrating particles. Hence frequency can be measured. Instruments like stroboscope can measure frequency of vibrating bodies. The seven swaras in musical scale (Table 1.1) are arranged in ascending order of pitch. Since pitch is not a measurable quantity, the list is in terms of frequency. The second harmonic of 256 Hz, namely, 512 Hz is identified by C' and it produces a sound sa'. Table 1.1 Musical scale S. No.

Note identification

Note

Frequency in Hz

1

C

sa

256

2

D

re

288

3

E

ga

320

4

F

ma

341

5

G

pa

384

6

A

dha

426

7

B

ni

480

1.3.2 Loudness Loudness of a musical sound is the degree of sensation produced by it on the ear of a listener. Loudness depends on the intensity of sound and sensitiveness of ear. Hence, it is a subjective quantity and cannot be measured by any instrument. Loudness is a characteristics of all sounds whether musical or nonmusical. Loudness of sound increases as intensity of sound increases but not proportionally. Even though loudness depends on intensity, both of them are different from each other. Whereas loudness cannot be measured, the intensity of sound can be measured. The intensity of sound is defined as the amount of sound energy that passes through unit area held normal to the direction of propagation of sound in one second. Thus intensity of sound is a physical quantity. Since loudness depends on the intensity of sound, naturally, it also depends on all those physical quantities on which intensity of sound depends. Following are the various physical quantities upon which the intensity and hence loudness of a sound depends.

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Acoustics and Structure of Solids

3

I. Amplitude of vibration of source It may be proved that the intensity (I) of a sound wave is given by I= 2p2a2n2lv …(1.1) where ‘a’ is the amplitude, ‘n’ is the frequency, ‘l’ is the density of medium and ‘v’ is the wave velocity. Hence the intensity of a sound wave varies directly as the square of the amplitude. (It may be felt in practice that when a given tuning fork is excited by two different forces, one after the other, and held near the ear, the listener will be able to hear a loud sound when the tuning fork is excited by a larger force than when excited by a smaller force. This is because larger force gives larger amplitude of vibration to the prong of tuning fork). It the amplitude of a sound wave is doubled, the intensity of sound increases to four times. Since loudness is a logarithmic function of intensity (Weber–Fechner law), it will also increase with increase in amplitude or greater is the amplitude of sound wave greater will be the loudness. II. Area of sounding body If the area of source of sound is large the loudness of sound produced will also be large. This can be understood easily from our everyday experience. The sound produced by an excited tuning fork is normally not audible unless the tuning fork is held very close to the ear of a listener. However, when an excited tuning fork is kept pressed on a sonometer, a loud sound is heard even when the listener is a little far away. This is because the sonometer becomes a source of sound due to forced vibrations of its particles. Since the area of a sonometer board is much larger than the area of a tuning fork, the loudness of sound produced is also large. III. Density of medium From Eqn. (1.1) it is clear that the intensity of a sound wave travelling in a medium is directly proportional to the density of medium. Hence greater is the density of medium in which sound is propagated greater is the intensity of sound. Consequently loudness of sound is also large. IV. Distance of listener It is known that inverse square law is obeyed in electrostatics, magnetism, gravitation and light. Sound also obeys inverse square law in the sense that as the distance of a listener from the source of sound is doubled, the intensity of sound at the position of listener is reduced to one fourth of the initial value. This may be understood from the following: Let P be the energy produced by a source of sound S in one second. If the surrounding medium is isotropic and the frictional resistance is regligible, the energy produced by the source of sound will get distributed equally in all directions. Let us imagine two listeners at A and B at distances d1 and d2 from the source (Fig. 1.1). Then the energy of sound, namely, P will be uniformly distributed over the areas 4πd12 and 4πd22 when sound waves from the source reach these two spheres (of radii d1 and d2 respectively with the source of sound at the centre) successively. Then the sound energy falling on unit area of each of the two spheres in one second are

P 4πd12

and

P 4πd22

respectively. Since energy per unit area per unit time is

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called the intensity of sound, it is clear that

P

Intensity of sound at A, I 1 =

4πd12 P

Intensity of sound at B, I 2 =

4πd22

Since P is a constant for a given source of sound, we have I1 µ I2 µ

1 d12 1 d 22

Thus the intensity of sound is inversely proportional to the square of the distance of listener from the source of sound.

r1 s

B

r2

Fig. 1.1 Inverse square law

V. Motion of medium If the medium present in between the source of sound and listener, say for example, air, moves towards the listener more number of waves would be carried to the listener in a given time than when the medium was still and hence the intensity and loudness of sound felt by the listener is more. In case the medium moves towards the source of sound, then the intensity and hence the loudness of sound felt will be less than what would be heard when the medium is not in motion. Thus motion of medium has an effect on both the intensity and loudness of sound.

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5

1.3.2.1 Weber–Fechner Law and Measurement of Loudness

As mentioned earlier loudness being the degree of sensation on the ear, it cannot be measured by any instrument. Weber–Fechner law in physiology states that the loudness (L) of a sound is proportional to the logarithm of intensity (I). Hence L µ log10 I L = k log10 I

or

…(1.2)

where k is the constant of proportionality. It should be understood that the loudness produced by a sound of intensity I, as given by Eqn. (1.2), cannot be measured. However difference in loudness, technically known as ‘intensity level’ is of importance and can be measured. Mathematically, intensity level (IL) is given by IL = L1 – L2

…(1.3)

where L1 and L2 are the loudness corresponding to two sounds of intensities, say I1 and I2. However, as is known, for any measurement there should be a zero level of reference. Hence L2 is replaced by the loudness corresponding to reference intensity, say, I0. The reference intensity I0 corresponds to the “threshold of audibility”. Threshold of audibility at any specified frequency is the minimum value of sound pressure of a pure tone of that frequency which is just audible. Its value is measured in newton per square metre. It has been experimentally found that at the threshold of audibility the sound of frequency 1000 Hz has a sound pressure of 2 × 10–5 N/m2 and an intensity of 10–12W/m2). Hence it is a convention that the reference intensity, I0, is taken as 10–12 N/m2 and reference pressure, P0, is taken as 2 × 10–5 N/m2. In this connection, it is also better that one knows about the “threshold of feeling”. The threshold of feeling at any frequency is the minimum value of the sound pressure of a pure tone that will cause a sensation of feeling or pain to the ear. The variations of the threshold of audibility and the threshold of feeling with respect to frequency are shown in the frequency–sensitivity curve in Fig. 1.2. 1000

Thresho ld of feeling

Sound pressure, N/m2

100 10 1 0.1 0.01 0.001

T hre

0.0001 0.00001 0

16

32

64

sho l d of audibility

128 256 512 1024 2048 4096 8192 16384 32768

Frequency, Hz

(Source: Acoustic–design and practice, R.L. Suri, Volume 1, Asia Publishing House, New Delhi, 1966) Fig. 1.2 Frequency-sensitivity curve

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Hence the intensity level (IL) is given by IL = L1 – L2 = k log10 I1 – k log10 I0 I  IL = k log10  1  …(1.4) I   0 where I0 = 10–12 W/m2, the intensity of sound of frequency 1000 Hz that corresponds to the threshold of audibility. From Eqn. (1.4), it is clear that the intensity level of a sound is defined as the logarithm of ratio of intensity of a given sound to the reference intensity, I0. From Eqn. (1.4), we can understand that the intensity level is a relative quantity and is not absolute. The unit of measurement of intensity level is decided by the value of the proportionality constant, k in Eqn. (1.4). The unit of intensity level is bel, when k = 1. The unit of intensity level is decibel, when k = 10. Therefore, for a given intensity of sound I, we can either write

(or)

or

 I  IL = log10   bel with k = 1 I   0

…(1.5)

 I IL = 10 log10    I0

…(1.6)

  decibel with k = 10  

To understand this, as an example, let us consider a sound of intensity 10–10 W/m2 and find its intensity level both in bel and decibel. Given intensity of sound, I = 10 –10 W/m2 Reference intensity, I 0 = 10 –12 W/m2 With k = 1,

 10 –10  IL = log10  –12  = log10 102  10   

= 2 log10 10 = 2 bel with k = 10,

 10 –10  IL = 10 log10  –12  = 10 log10 102  10 

= 2 × 10 log10 10 = 20 decibel or 20 dB Thus a sound of intensity 10–10 W/m2 produces an intensity level of 2 bel which is the same as 20 decibel or we also understand that 1 bel is equal to 10 decibel. We shall now try to understand the meaning of zero level of reference for intensity level. You are aware that the threshold of audibility, namely, an intensity of 10–12 W/m2 is taken as reference intensity. What would be the intensity level produced by a sound of such intensity? Given intensity of sound, I = 10 –12 W/m2 Reference intensity, I 0 = 10 –12 W/m2 \ Intensity level

 I IL = 10 log10   I0

   

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7

 10 –12  = 10 log10  –12   10 

= 10 log101 = 0 dB Thus a sound of intensity corresponding to threshold of audibility produces an intensity level of zero decibel or in other words a just audible sound can produce an intensity level of zero decibel. It is also worth mentioning here that an intensity level of 120 dB causes threshold of feeling i.e., an intensity level of 120 dB gives pain to the ear. It is also important to know that we should not add two intensity levels numerically to get the resultant intensity level. (See worked Examples 1.14 and 1.15) At this stage, it is better to recollect that loudness cannot be measured and intensity level can be measured. However, in practice, very often intensity level and loudness are used interchangeably i.e., the term “loudness” is used instead of “intensity level” very frequently. It will be quite interesting to see a few more examples that makes one understand the intensity level, before we discuss “phon” in Section 1.3.2.2. Example 1.1 Calculate the intensity of sound that produces an intensity level of 1 dB. Solution

\

 I IL = 10 log10    I0  I 1 = 10 log10   I0

  dB      

 I  0.1 = log10    I0  I antilog 0.1 = I0 I = (antilog 0.1) × I0 = 1.26 × I0 = 1.26 × 10–12 W/m2 (\ I0 = 10–12 W/m2)

Ans. A sound of intensity 1.26 × 10–12 W/m2 can produce an intensity level of 1dB. Example 1.2 By how much should the intensity of a sound increase to produce an increase in intensity level (or) loudness by 1dB? Solution Given IL2 – IL1= 1 dB \

I  1 = 10 log10  2  – 10 log10  I0 

I  1 I  0

   

8


where I1 and I2 are the initial and final intensity of sound respectively such that the increase in intensity level is 1dB. \

 I 1 = 10 log10  2   I0  I = 10 log10  2   I1

 I    – log10  1    I 0   I0 I0

   

 I I  = 10 log10  2 × 0   I 0 I1   I  1 = 10 log10  2   I1 

or

I  0.1 = log10  2   I1 

antilog 0.1 = \

I2 I1

I 2 = (antilog 0.1) × I1 I 2 = 1.26 I1

Increase in intensity that   \ produces an increase in  intensity level by 1 dB 

= = = =

I2 – I1 1.26 I1 – I1 (1.26 – 1) I1 0.26 I1 or 26% of I1 Hence, when the intensity of sound increases by 26%, the intensity level increases by 1dB. Example 1.3 What is the intensity of sound that can produce an intensity level of 50 dB? Solution

 I IL = 10 log10   I0

   

 I 50 = 10 log10  I  0  I  5 = log10    I0 

   

I


Antilog 5 = \

9

I I0

I = (antilog 5) × I0 = 105 × 10–12 = 10–7 W/m2

Ans. A sound of intensity 10–7 W/m2 produces an intensity level of 50 dB. Example 1.4 The intensity level of sound at a given place is found to be 50 dB. By how much one should increase the intensity of sound to get an intensity level of 51 dB? Solution Let I1 be the initial intensity of sound corresponding to IL of 50 dB, represented by IL1. Let I2 be the final intensity of sound corresponding to IL of 51 dB, represented by IL2. \

I 50 = 10 log10  1  I0

(or)

I 1 = 10–7 W/m2 (Refer example 1.3)

and

\

\

   

I  51 = 10 log10  2  I   0  I2  5.1 = log10    I0  I2 antilog 5.1 = I0 I 2 = (antilog 5.1) × I0 = 1.26 × 105 I0 = 1.26 × 105 × 10–12 = 1.26 × 10–7 W/m2 = 1.26 I1 Increase in intensity = I2 – I1 = 1.26 I1 – I1 = (1.26 – 1) I1 = 0.26 I1 = 26% of I1

Ans. When the intensity of sound increases by 26% of initial intensity, the IL increases by 1 dB, say from 50 dB to 51 dB. (Refer example 1.2 also) 1.3.2.2 Phon

We have seen earlier that a sound of frequency 1000 Hz with an intensity of 10–12 W/m2 is just audible. Hence 10–12 W/m2 was taken as the reference intensity when intensity level was measured either in bel

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or decibel. However, it has been proved that a sound of frequency 30 Hz with an equal intensity, namely, 10–12 W/m2, is not at all audible. A sound of frequency 30 Hz becomes audible only when its intensity is increased to 10–6 W/m2. This makes us understand that sounds of different frequencies but having same intensity will not produce equal loudness. This is due to (i) variation in sensitiveness of ear for different individuals, and (ii) the fact that limits of audibility depend both on intensity and frequency. Hence scientists have adopted a standard source of frequency 1000 Hz with which all sounds (of different frequencies) are compared. This leads to another unit for measuring intensity level (or) loudness level and is called ‘phon’. Phon is also said to be the unit of “equivalent loudness”. The loudness level of a sound of a given frequency in phon is equal to the loudness in decibels of an equally loud pure tone of frequency 1000 Hz as heard by a normal observer standing in front of the source. To understand the meaning of phon just now explained, we shall see the method of estimating the loudness level in phon of a given sound of certain frequency: Let us assume that we have to find the loudness level of a tone, say, of frequency 512 Hz. Now a normal observer should perform an experiment with the given source of sound and a standard sound source of frequency 1000 Hz as follows. The normal observer should now hear with both ears the sounds produced by the two sources alternately and keep adjusting the loudness of sound produced by the standard source of frequency 1000 Hz until the normal observer is able to judge that the sound produced by 1000 Hz standard source is as loud as the sound produced by the given source of frequency 512 Hz. At this stage the normal observer will measure the intensity of sound of the standard source 1000 Hz and calculate its intensity level in decibels with respect to the reference intensity of 10–12 W/m2. If the intensity level of the standard source of frequency 1000 Hz at this stage is ‘n’ decibels, then the given source of sound of frequency 512 Hz is said to have an equivalent loudness or loudness level of ‘n’ phon. 1.3.2.3 Sone

Loudness level is also measured by another unit, namely, sone. One sone is defined as the loudness experienced by a typical listener when listening with both ears to a tone of frequency 1000 Hz at a level of 40 dB above threshold. The relation between phon and sone as derived by Stevens is log S = 0.03P – 1.2 …(1.7) where S is loudness in sone and P is the loudness in phon. 1.3.2.4 Sound Pressure Level (SPL)

We know that sound waves exert pressure variations in a medium in which they travel. Since the voltage outputs of microphones and hydrophones used in acoustic measurement are proportional to the acoustic pressure, the acoustic pressure is the most readily measurable variable in a sound field. Hence, nowadays, sound pressure levels are more widely used in specifying sound levels than the intensity levels. Hence we will sea an expression for sound pressure level. It may be proved that if r0 is the constant equilibrium density of a medium in which a plane progressive wave travels with a velocity v, then the intensity of the plane progressive wave is given by

Pe2 …(1.8) 2ρ 0 v where Pe is the effective pressure of root mean square pressure of plane progressive wave. Hence I=

I


11

it is clear that instead of intensity you can use the “effective pressure” in calculations. Corresponding to the intensity of threshold of audibility i.e., I0 = 10–12 W/m2, we also have the effective pressure corresponding to threshold of audibility, namely, P0 = 2 × 10–5 N/m2 (or) in other words, 2.×10–5 N/m2 should be used as reference pressure for a plane progressive wave travelling in air medium. For convenience we will use “P” in the place of “Pe”, since very often in practice “effective pressure” is simply mentioned as “pressure”. Therefore, now, we can write, for a sound wave of intensity I,

P2 2l 0 v For a sound wave of reference intensity I0, I=

…(1.9)

P02 I0 = 2l 0 v Dividing Eqn. (1.9) by Eqn. (1.10), we get I P2  P = 2 =  I0 P0  P0

   

…(1.10)

2

…(1.11)

Using Eqn. (1.11), we can now get an expression for sound pressure level as an alternative for  I  intensity level of sound as given below, by simply replacing   in Eqn. (1.6) by  I0  that

Now,

 I Intensity level = 10 log10   I0

  decibel.  

2

P   . We know    P0 

…(1.6)

2

 P Sound pressure level = 10 log10   decibel    P0 

\

 P = 20 log10   P0

  decibel  

 P Sound pressure level = 20 log10    P0

  decibel  

…(1.12)

While using Eqn. (1.12) for calculating sound pressure level, the value of the reference effective pressure P0 should be taken as 2 × 10–5 N/m2 for air. It is worth mentioning here that “sound pressure level” is easily measured than “intensity level” in actual practice and hence in use.

1.3.3 Quality or Timbre Quality is one of the characteristics of a musical sound. It helps one to differentiate between two sounds of same pitch and loudness played on two different musical instruments or produced by two different persons.

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When same song is played by veena and flute simultaneously, you are able to differentiate between the two sounds. It is because of the difference in quality of sound produced by them. The wave forms produced by the two instruments are different. A tuning fork can produce a pure note i.e., sound of one particular frequency. However, the musical notes or sounds produced by different musical instruments have both the fundamental frequency, say n, and other tones of higher frequencies called overtones, namely, 2n, 3n, 4n etc. Of course, not necessarily all the overtones have to be present. The resultant sound of all these frequencies decides the quality of sound. Hence the quality of sound depends on (i) the number of overtones produced along with the fundamental frequency, and (ii) the relative intensity of overtones. We know that closed end organ pipes are capable of producing overtones whose frequencies are odd number multiples of the fundamental. However, open end organ pipes produce overtones whose frequencies are integral multiples (i.e., both odd number times and even number times) of the fundamental frequency. Hence it is clear that sounds produced by open end organ pipes are of better quality than the sounds produced by closed end organ pipes.

1.4 ACOUSTICS OF BUILDINGS Until the beginning of twentieth century, people did not give any importance to the acoustic characteristics of buildings, lecture halls and auditoria. Hence many a times, the lectures, musical performance etc., given inside them were not that much clear. When Harvard University constructed an auditorium (Fogg Art Museum Hall) during the year 1900, many of the lecture programmes and orchestra performed inside the auditorium were miserably poor acoustically and nothing could be understood. Wallace C. Sabine, who was the professor of Physics in Harvard University was requested to identify the cause of the problem and suggest a suitable remedy. It was at that time research towards acoustical characteristics of buildings started. The research work of Sabine provided good results. For any auditorium, lecture halls or buildings to be acoustically good, Sabine suggested the following requirements: (i) The sound must be sufficiently loud everywhere inside the hall. (ii) The quality of sound produced by a source of sound (like human beings and instruments) must remain unaltered. (iii) The successive sounds of speech and music must remain distinct (i.e., must not overlap) and free from extraneous noises (iv) There should be no echoes more than necessary for the upkeep of continuity (v) There should not be undersirable focussing of sound in any part of the hall and depletion in any other part. (vi) There should not be unpleasant reinforcement of and articulation by objects inside the hall. Depending on the purpose for which a hall or builiding is designed, some or all of the above conditions are to be satisfied to be acoustically good. To achieve the above requirements so as to make any hall or building to be acoustically good you have to consider the effects of the following parameters and make suitable efforts: 1. Reverberation 2. Loudness of sound 3. Focussing effects

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4. Echelon effect 5. Noises 6. Resonance 7. Interference 8. Echoes We shall discuss all these parameters.

1.4.1 Reverberation When a sound is produced inside a hall, it spreads out in all directions and is expected to be heard once by an audience inside the hall. But, inside the hall, the sound waves get reflected so many times and a series of waves of successively decreasing amplitudes pass through the listeners’ ears until they die due to frictional forces in the air. As a result, even though the source of sound produced a particular sound and stopped, the sound is continuously heard for a short interval of time, until the intensity falls below the limit of audibility. It was found by Sabine, in his research, that each sound wave undergoes 200 to 300 reflections inside a hall before it becomes inaudible. This phenomenon of persistence of sound inside a hall for some time even after the source of sound is stopped is called reverberation. Using an organ pipe of frequency 512 Hz ,Sabine found the time for which each sound wave is heard inside the hall even after the sound source is stopped. Sabine’s research showed that every sound wave became inaudible when its intensity fell to one-millionth of its intensity just when stopping the organ pipe. Hence reverberation time is defined as the time taken by sound to fall to one-millionth of its intensity from its intensity just when the source is stopped. If we talk in terms of decibel, the reverberation time is defined as the time taken by sound to fall in its intensity level by 60 dB. (Please refer note at the bottom of the page). Sabine also derived a formula for reverberation time of a hall and is as given below: T=

0.16 V

…(1.13)

i=n

∑ α ds i

i

i =1

Note : Let the intensity of sound just when it is stopped be I. Just when the sound becomes inaudible for the listener, the I intensity of sound is 6 i.e., one-millionth of intensity just when it is stopped. 10 \ Fall in intensity level = Initial IL – Final IL

 I 106   I   = 10 log10   –10 log10   I0   I0     I   I  = 10 log10   – 10 log10  6   I0   10 I 0   I 106 I  0 × I   I 0  = 10 log10 106 = 60 log10 10 = 60 dB

= 10 log10 

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where V is the volume of hall in m3, ai is the absorption coefficient of ith material/surface, dsi is the area of ith absorbing material/surface and n is the total number of absorbing items in the hall. (The derivation of Sabine's formula is given in Section 1.5). The factors which can help us control reverberation time in a hall and their effects are discussed below: I. Absorption Reverberation time depends on the absorption coefficient of various surfaces available in a hall such as cushions, carpets, curtain etc. Greater the absorption shorter will be the reverberation time, and vice versa. (Definition of absorption coefficient and measurement of absorption coefficient are discussed in Sections 1.6 and 1.6.1 respectively). II. Reflection Reverberation time depends on the reflection property of walls, floor, ceiling etc. If the materials inside the hall are good reflectors, they reflect greater portion of the incident intensity (or in other words loss in intensity is less) and hence it takes longer time for the sound to become inaudible. Therefore longer would be the reverberation time. On the other hand if the materials inside the hall are not good reflectors, shorter would be the reverberation time. III. Frequency of sound It has been found that for most of the materials the absorption coefficient increases with increase in frequency of sound. Hence, the reverberation time is shorter for high frequency notes, in general. IV. Intensity of sound just when sound source is stopped It is a simple logic that sounds of higher intensity would take a longer time to get reduced in their intensity to inaudible level. Hence, greater the intensity of sound just when it is stopped, longer would be the reverberation time. V. Volume of hall The reverberation time also depends on the volume of the hall. According to Sabine, T=

0.16 V i= n

∑ α ds i

…(1.13)

i

i =1

where V is the volume of the hall. Thus we realise that reverberation time is one of the most important parameters that can decide the acoustic characteristics of a hall. It is but natural to put a question, “what should be the optimum reverberation time of a hall?” There cannot be a single answer for this question. It is because the reverberation time of a hall is decided for the purpose for which it is built. The reverberation time of a lecture hall, a theatre and a concert hall need not necessarily be the same. For concert halls of different sizes, the reverberation time need not be equal since the reverberation time depends on the volume of the hall and the type of sound (speech of music) in the hall. In general, for most of the listeners a reverberation time of 0.5 s is found to be good for speech and it is in between 1 and 2 s for music. A reverberation time between 1.1 s and 1.5 s is quite good for a hall of capacity, approximately 1700 m3. However for a bigger hall of volume between 7000 m3 and 18,000 m3, the optimum reverberation time is about 2.3 s. It is very important to understand that the reverberation time should be the same for a hall irrespective of the fact that the hall is empty or full

I


15

with audience. This may be accomplished easily by using suitable absorbing materials on the seats which will have nearly the same absorption coefficient of an audience (while absorption effect of all materials remaining the same for a given type of sound in the hall it is the audience who sits on the seat can alter or change the reverberation time to a certain extent. The audience may be a child, an adult, a man or a woman who will have slightly differing obsorption property. When the audience sit on the seat, he or she absorbs certain amount of sound energy. When the seat is vacant, the seat should absorb the same amount of sound energy as that would be absorbed by an audience. Then only the reverberation time of the hall will be same whether the hall is empty or full with audience. By now, you might have understood that if the reverberation time of a hall or auditorium is not upto the expected standard optimum value, the acoustical condition of the hall may be improved either by introducing or by removing estimated amount of absorbent surfaces (To understand this see Examples 1.8 and 1.9).

1.4.2 Loudness of Sound Every audience inside a hall, wherever he or she sits, should be able to enjoy the performance on the stage or be able to listen clearly any lecture/speech delivered on the stage. Then only the hall is said to be acoustically good. To achieve this the sound must be sufficiently loud everywhere inside the hall. The loudness of sound inside the hall can be made uniform by properly utilising the reflection properties of walls, floor, ceiling etc. By using public addressing system which consists of microphones, amplifiers and loud-speakers, the loudness of sound inside the auditorium may be increased and thereby every listener can be made to hear the sound comfortably without strain. Very often, sounding boards kept on the stage reflect the sound towards audience and increases the loudness. Plane reflecting surfaces help to increase the loudness.

1.4.3 Focussing Effects Any regular curved surface inside an auditorium will be capable of focussing sound waves in a line and thereby provide uneven distribution of sound. This should be avoided. Hence curved surfaces should be suitably designed. Experiments have shown that curved ceiling which has a radius of curvature equal to twice the height of the hall is useful. Such a ceiling avoids focussing of sound in a row or at a point inside the hall.

1.4.4 Echelon Effect (or) Musical Echo You know that many halls will have a number of steps arranged throughout the length of the hall by the sides of regular rows of seats (Fig. 1.3). Similarly regularly spaced designs, called palings can be seen on the ceiling (Fig. 1.4). The number of equidistant and parallel steps/palings form a grating-like system. When a single sound pulse falls on the system, it is broken up into a number of pulses by successive reflections from the steps/palings. Since these steps/palings are equidistant from one another, the portion of the wave striking each step/paling and getting reflected there travels a little farther and arrives a little later at the listeners’ ear, than the portion striking the previous step/paling. The succession of pulses at regular intervals is heard as a short musical note of definite frequency. This is called echelon effect. If d is the distance between two successive steps/palings, v, the speed of sound and q, the angle of deviation of the

16


reflected pulses, the period of the echo i.e., the time interval between two pulses will be Hence the frequency of the musical echo will be

2d cos θ . v

v . For normal incidence, the frequency will be 2d cos θ

v . For example, if the width of each step is 0.5 m (Fig. 1.4) and if the speed of sound at room 2d 350 temperature (30ºC) is taken as 350 m/s, the frequency of musical echo would be = 350 Hz, 2 × 0.5 assuming normal incidence (q = 0º). Hence the audience will hear a musical echo of frequency 350 Hz along with the music or speech from the stage. Naturally, this will affect the acoustic quality of the hall. This echelon effect can be overcome by spreading a carpet over the steps and/or by making the spacing of steps irregular. Similarly by making the spacing of palings irregular on ceiling, echelon effect can be minimised.

0.5 m

d

Fig. 1.3 Flight of steps

d

θ

θ osθ dc

Fig. 1.4 Palings of ceilings

1.4.5 Noises Very often sounds produced by moving automobiles (like two wheelers, cars and even sometimes aeroplanes etc.) are heard inside the auditorium or hall. Such a disturbance is called ‘airborne’ disturbance. This can be avoided or minimised by reducing the number of doors and windows of the hall (or) all the

I


17

doors and windows of the hall may be closed when a performance takes place. But this will result in either less light or no light inside the hall. Of course air circulation and ventilation also become a problem. This can be definitely overcome by airconditioning the auditorium/hall and by arranging sufficient number of light fittings inside. If airconditioning is not possible, ventilating fans and less number of doors and windows may be used. Similarly when water flows through drain pipes fixed outside the hall, a disturbing sound can be heard inside the hall. This ill effect can be minimised by using rubber gasket junctions in the pipe.

1.4.6 Resonance Experiments have proved that natural frequency or resonant frequency of a hall is inversely proportional to the square root of the volume (V) of the hall i.e., resonant frequency µ

1 V

Since the volume of a hall would be normally large, the resonant frequency would be small. Very often it will be below the audible limit and hence the effect of resonant frequency is not felt. However, the resonant vibrations become noticeable when the source of sound is very powerful. These sources of sound will produce, due to resonant effect, unpleasant sensation to the listener. But if resonant surfaces like panelling etc., are suitably covered, the ill effect of resonant vibrations gets reduced. Loose hanging pendants and latches, hollows and crevices would pick up sounds of certain frequencies and produce bad resonance effects. While it is not always possible to avoid the use of such objects, special care must be taken that their effects are minimal.

1.4.7 Interference When a sound of steady and constant frequency is produced, interference effects are produced in halls leading to maxima and minima of sound at various places. The variation of intensity in the room can be detected by means of microphones moved about inside the room. Different reponse can be felt at different positions in the room, even in the absence of focussing surfaces. The ill effect of interference may be minimised by avoiding smooth or polished corners in the room.

1.4.8 Echoes In large lecture halls, echoes result from sound getting reflected at walls and surfaces at great distances returning long after original sound has died. They can be eliminated almost entirely by making the surfaces of walls rough. When the hall is full with audience, the effect of echoes will not be felt. Similarly when a hall has large number of windows and doors, echoes are practically absent. In music halls, however, the presence of faint echoes help in improving musical effects, when the ceilings of the hall are polished and low. When the ceilings are high, the sounds get damped before they get reflected. Thus by controlling various factors (discussed through sections 1.4.1 to 1.4.8) governing acoustical conditions of a hall, acoustically good lecture halls and concert halls can be built. This subject of architectural acoustics or acoustics of buildings is growing ever since Prof. Sabine started this in 1900 and more and more technological advancements are made and in this century we have very many acoustically good halls, auditoria and theatres spread throughout the world.

18


1.5 SABINE’S FORMULA FOR REVERBERATION TIME Sabine’s formula for reverberation time is derived here based on the assumption that the sound energy is distributed uniformly in all directions inside a hall. The derivation is done in three steps. The aim of the three steps are: Step 1: To calculate the rate at which energy is incident on walls and other surfaces in terms of the average energy density, E and hence the rate at which energy is absorbed by the wall. Step 2: To calculate the value of E in terms of the rate of emission of energy or power P of the sound source and also to calculate the rate at which energy decreases when sound sources are switched off. Step 3: To frame and solve a differential equation for finding the reverberation time based on the principle that the rate of change of E is equal to the excess of P over the rate at which energy is absorbed. Let us now derive an expression for the reverberation time of a hall.

Step 1 Let ds be a small element of area on the wall. Let O be its centre. Let ON be a normal drawn to the area ds. With O as centre, let us draw two semicircles of radii r and r+dr in a plane containing the normal ON. We shall now consider two radii OA and OD at angles q and q + dq from the normal. Thus we consider a small area ABCD at a radial distance r from the area ds (Figure 1.5). Here AB = dr and AD = rdq from the geometrical considerations. \ Area ABCD = (rdq) × dr Also EA = r sin q dS N

B r sinθ

E

r θ

Area dS

A

dr C D

dθ

O

Fig. 1.5 Area element used for deriving rate of absorption of sound

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19

Let us now rotate this area ABCD about the normal ON through an angle df. This means that the area ABCD will move through a distance (r sin q) df and hence we get a small elemental volume dV (given by ABCDA'B'C'D') equal to (rdq) {(r sinq) df }(Figure 1.6). \ dV = r2dr sin q dq df …(1.14) dS N

r sin θdφ A' dr

dφ E

r sinθ

θ

Area dS

B'

r dθ

A

B

Volume dV C'

D' C D

rdθ

O

Fig. 1.6 Volume for surface element used in deriving rate of absorption of sound

If we assume that a performance is going on inside the hall, then the sound energy starts from the stage and travels uniformly in all directions (according to the assumption made). If this is the case, let E be the average energy density in the hall i.e., E is the average energy available in unit volume of the hall (Refer Note 1)

Average energy present  = (r2 dr sinq dq df) E \ in volume dv 

…(1.15)

Since we have assumed that energy travels equally in all directions from the stage,

Er 2 dr sin θ dθ dφ Average energy travelling through =  4π unit solid angle in any direction 

…(1.16)

(We know that the solid angle subtended at any point inside a closed volume is 4p)

Note 1: During any performance the energy produced by a source of sound is continuously changing. Further all sources of sound are not producing equal amounts of energy at any given instant of time. Therefore the total energy produced by all sources of sound is not a constant. Therefore we should consider the average energy produced by all sound sources. Keeping in mind halls of different volumes, we consider average energy density, E i.e., average energy in unit volume of a hall.

20


But the solid angle subtended  ds cos θ  by the area ds at the elemental = r2  volume dV 

=

Eds sin θ cos θ dθ dφ dr 4π

…(1.17)

If v is the speed of sound at room temperature (Refer Note 2), Average energy   that falls on area  = ds in one second 

θ=

π 2

φ =2 π r = v

θ=0

φ=0

θ=

=

Eds sin θ cosθ dθ dφ dr 4π r =0

∫ ∫ ∫ π 2

Eds sinθ cosθ dθ 4π θ=0

∫

θ=

Eds ×V = 4π

∫

φ=0

…(1.18)

r =0

φ= 2 π

∫ dφ

φ=0

θ= 0

π 2

EVds × 2π sin θ cosθ dθ 4π θ= 0

∫

π 2

EVds 1 2 sinθ cosθ dθ × 2 2 θ= 0

∫

θ=

=

∫ ∫ dr

sin θ cos θ dθ

θ=

=

r =v

dφ

π 2

θ=

=

φ= 2 π

π 2

EVds sin 2θ dθ 4 θ= 0

∫

θ=

π

EVds  cos 2θ  2 –  = 4  2  θ =0

Note 2: We should understand that Eqn. (1.18) gives the amount of average energy that travels towards the area ds from that present in a volume dV, irrespective of the distance of dV from the area ds. If v is the speed of sound at room temperature, then we can understand that the average energy present in all elemental volumes dV which are present upto a maximum distance of v from the area ds can alone reach the area ds in one second because v is equal to the distance through which sound can travel in one second. Similarly average energy present in all the elemental volumes dV which lie within the range π and f = 0 to f = 2p can alone reach the area ds. q = 0 to q = 2

I


\

21

=

EVds ×1 4

=

EVds 4

EVds Average energy that falls  = 4 on area ds in one second 

…(1.19)

If a is the absorption coefficient of the material of area ds of the wall, then

average energy absorbed EVds  = ×α by area ds in one second  4 \

…(1.20)

Average energy absorbed by i =n EVdsi  all surfaces ( Refer Note 3) in = × αi 4  one second i =1

∑

where µi is the absorption coefficient of ith absorbing material, dsi is the area of ith absorbing material and n stands for the total number of absorbing materials. Hence

EV average energy absorbed by ×  = 4 all surfaces in one second  =

i=n

∑ α ds i

i

…(1.21)

i =1

EVA 4

…(1.22)

i=n

where

A =

∑ α ds i

i

…(1.23)

i =1

Step 2 In step 1, the average energy density inside the hall was assumed to be E. Now, we shall derive an expression for E. If P is the average power of the source of sound i.e., average energy produced by the source in one second, then the maximum average energy that can be absorbed by all surfaces in one second must be equal to this, i.e., EmVA P= (from Eqn. 1.22) …(1.24) 4 where Em is the maximum value of average energy density that can be attained inside the hall.

Note 3: We know that every area ds of all surfaces like wall, floor, ceiling, curtain etc. are not of same material i.e., different areas ds of absorbing surfaces have different absorption coefficients. Hence the absorption by all surfaces has to be calculated by summation process.

22


From Eqn. (1.24), we have 4P …(1.25) VA Now at any instant of time E, is the average energy density and assuming V to be the volume of the hall, we can write

Em =

Average energy present inside  the hall at a given instant of   = EV time when the average energy   density is E

…(1.26)

Hence rate of growth of   d average energy density i.e.,   = dt (EV ) average energy available in  one second dE …(1.27) dt This should be equal to the difference between average energy produced by sources of sound in one second and the average energy absorbed by all surfaces inside the hall in one second. i.e.,

= V

Rate of Rate of absorption Rate of growth of average energy = production of average – in the hall of average sound energy

\

V

EvA dE = P– dt 4

or

P  vA  dE – E  = V  4V  dt

…(1.28)

Let

vA = β 4V

…(1.29)

\ Equation (1.28) becomes dE  4β  =   P – Eβ dt  vA   4P  dE β + Eβ =   vA  dt Multiplying both sides by ebt, we get   dE  4 P  βt + Eβ eβ t =  β e  dt    vA 

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23

d  4 P  βt ( Eeβt ) =  βe dt  vA  Integrating both sides, we get

or

 4P  βt e + K Eeβt =   vA 

…(1.30)

where K is the constant of integration (Refer Note 4). I. Growth of average energy density From Eqn. (1.30), we can derive and understand the way the average energy density grows and reaches its maximum value inside the hall. Just when a sound source is switched on, t = 0 and E = 0 \ Equation (1.30) becomes 4P +K 0= vA 4P K= – …(1.31) vA Substituting this value of K in Eqn. (1.30), we get  4 P  βt 4P e – Eeβt =  vA  vA  4 P 4 P –β t E= e – vA vA

=

(

4p – e –βt vA

)

= Em (1 – e –βt ) (from Eqn.1.25) \

E = Em (1 – e –βt )

…(1.32)

From Eqn. (1.32), we understand that as t increases, e–bt decreases and hence E increases. A graph drawn between time, t and the average energy density, E is exponential as shown in Figure 1.7. From Eqn. (1.32), we also have at t = ¥, E = Em. In Eqn. (1.32), E and Em represent the average energy density at any instant and maximum value of average energy density inside the hall respectively (Refer Note 5).

Note 4: From Eqn. (1.30), we can get expression for average energy density, E. However, we have to consider two cases. In the first case the situation is that we try to see the nature of E from the moment the source of sound is switched on and until reaches its maximum value i.e., the growth of average energy density value from its zero value (the nature of E as the sound energy increases from zero value). In the second case the situation is that we try to see the nature of E from the moment the sound source is switched off and until it reaches its zero value i.e., the decay of average energy density (the nature of E as the sound energy decreases from its maximum average value to zero value). Note 5: Hence for a hall of any volume, E and Em will represent the energy available inside the hall at any instant and the maximum value of energy inside the hall respectively.

24


Average energy density, E

Em

time, t

O

Fig. 1.7 Growth of average energy density

II. Decay of average energy density From Eqn. (1.30), we can also understand the way the average energy density inside the hall decreases from its maximum value and reaches zero value after the sound source is stopped or switched off. Let us assume that when the average energy density inside the hall is at its maximum value, namely, Em, the sound source is stopped/switched off with a timer being started simultaneously. Then at t = 0, E = Em and P = 0. Hence Eqn.(1.30) becomes Em = 0 + K K = Em …(1.33) Substituting this value of K in Eqn. (1.30), we get Eebt = Em (\ P = 0) E = Eme –bt

…(1.34)

In Eqn. (1.34), as t increases e–bt decreases and hence E decreases exponentially. Hence, after the sound source is stopped/switched off, the average energy density inside the hall starts decreasing exponentially (Figure 1.8). From Eqn. (1.34), we also get at t = ¥, E = 0 (Refer Note 6).

Step 3 In step 2, we have seen the way the average energy density decreases after the source of sound is stopped/switched off. This helps us to derive an expression for the reverberation time of a hall. We know that the reverberation time of a hall is defined as the time taken by sound to fall to one-millionth of its intensity from its intensity just when the source is stopped. It may be proved that reverberation Note 6: Based on a similar logic explained under the growth of average energy density, we can also understand that, for a hall of any volume, after the sound source is stopped/switched off, E represents the energy and it keeps decreasing exponentially and finally reaches zero value.

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25

time may also be defined as the time taken by the sound to fall in its average energy density value to one-millionth of its value from the moment the source is stopped (Refer Note 7).

Average energy density, E

Em

O

time, t

Fig. 1.8 Decay of average energy density

Hence in Eqn. (1.34), when t is replaced by T, the reverberation time, E, should be replaced by

Em 106

. \ Equation (1.34) becomes

Em 106

= Em e –β T

1

= e –βT 106 Taking reciprocal on both sides 106 = ebT Note 7: We know that, intensity of sound =

energy area × time

\ For a given area of absorbing surfaces/materials, by the time the intensity falls to one-millionth of its initial value, the energy also falls to one-millionth of its initial value. Further we know that, energy density =

energy volume

\ In a hall of given volume, by the time the energy decreases to one-millionth of its initial value, the energy density also decreases to one-millionth of its initial value. This is true for the value of average energy density also. \ Reverberation time may also be defined as the time taken by the sound to fall in its average energy density value to one-millionth of its value from that at just when the source is stopped.

26


Taking natural logarithm on both sides, we get 6 loge10 = bT (or) bT = 6 loge10 = 6 × 2.303 Substituting for b from Eqn. (1.29), we have vA × T = 6 × 2.303 4V

\

T=

6 × 2.303 × 4V vA i=n

Assuming v = 350 m/s at room temperature and replacing A by

∑ α ds , we get i

i

i =1

Reverberation time,

T=

0.16V i=n

∑ α ds i

…(1.35)

i

i =1

This is the Sabine’s formula for reverberation time.

1.5.1 Measurement of Reverberation Time There are a number of methods like stationary wave method, reverberation chamber method etc. to measure the reverberation time of a hall. Some of these methods are subjective methods which depends on the judgements made by observers and hence may not be accurate methods. Some of them are objective methods which do not depend on judgements made by observers and hence are better than subjective methods. One of the simplest objective methods for measuring the reverberation time is explained below. A high speed sound level recorder which has a calibrated microphone, amplifier and rectifier is used inside the hall whose reverberation time is to be found. Even when all the sources of sound inside the hall are ‘ON’, the recorder is switched ‘ON’ and the sound intensity level inside the hall is continuously recorded by a stylus on a moving strip of paper (time on X-axis and sound intensity level on Y-axis). When all the sound sources are stopped/switched off, the sound intensity level inside the hall automatically starts decreasing continuously. It shall be seen that the graph relating the time and sound intensity level becomes a straight line parallel to the time axis “as and when” the found intensity level becomes inaudible i.e., when the found intensity levels has just fallen to one millionth of its intensity from just when the sources of sound are stopped/switched OFF (for detecting which the microphone is previously calibrated). The difference in time between “the time at which the sources of sound are stopped/ switched OFF” and “the time from which almost a straight line starts getting recorded” gives the reverberation time of the hall. Sound intensity level recorders which are capable of measuring reverberation time of “0.2s accuracy and more” are available nowadays. A typical output of sound intensity level recorder is shown in the Figure 1.9.


27

Intensity level, dB

I

0

1.0

2.0

3.0

time, t Fig. 1.9 Time-sound intensity level curve

1.6 ABSORPTION COEFFICIENT Like any other physical quantity, the measurement of absorption coefficient also requires a ‘standard’ and ‘unit’. This will help measurement of absorption by different materials/surfaces and have a comparison among them. It is well known that all sound waves which reach an open window will pass through the window and the sound will not be available for any audience. Hence ‘open window’ is considered to be the best absorber of sound and unit area of open window is taken as reference for defining the absorption coefficient. The absorption coefficient (a) of a given surface/material is defined as “the reciprocal of its area which absorbs the same sound energy as that absorbed by unit area of an open window”. OR The absorption coefficient (a) of a given surface/material may also be defined as “the ratio of the sound energy absorbed by the given area of the surface/material in a certain time to the sound energy absorbed by an equal area of a perfect absorber such as an open window in an equal time.” OR The absorption coefficient (a) of a given surface/material is also defined as “the ratio of amount of sound energy absorbed by the area of given surface/material in a certain time to the sound energy incident on the area of given surface/material in an equal time”. \

Absorption coefficient of ü a = a given surface/material ýþ

1 Area of given surface/material which will absorb same amount of energy absorbed by unit area of open window

OR Absorption coefficient of ü a = a given surface/material ýþ

Sound energy absorbed by given area of surface/material in certain time Sound energy absorbed by equal area of open window in equal time

28


Absorption coefficient of ü a = a given surface/material ýþ

Sound energy absorbed by given area of surface/material in certain time Sound energy incident on the given area of surface material in an equal time

The unit of absorption coefficient is “open window unit”, abbreviated as O.W.U. “Sabine” is also used in the place of “O.W.U.” as a respect to him.

1.6.1 Determination of Absorption Coefficient There are many methods by which absorption coefficient of a surface/material can be measured. We shall discuss here three methods. Method I.

The absorption coefficient of a material can be found by estimating the reverberation time of a hall without and with the given material inside the hall. Let T1 be the reverberation time of a hall without the given material in it. Then T1 =

0.16V

…(1.35)

i=n

∑ α ds i

i

i =1

i=n

where

∑ α ds i

i

represents the total absorption inside the hall without the material of our interest.

i =1

Now let us introduce a known area S of the given material whose absorption coefficient, a is to be determined. Let T2 be the reverberation time of the hall with the given material inside. Then T2 =

0.16V

…(1.36)

i=n

∑ α ds + α s i

i

i =1

Now taking reciprocals of 1.36 and 1.35 and finding the difference between them, we have i=n

1 1 = – T2 T1 =

\

∑

i=n

∑ α ds

α i dsi + α s

i =1

0.16 V

i

–

i

i =1

0.16V

αs 0.16 V

0.16V Absorption coefficient of ü a = a given material/surface ýþ S

1 1  –   T2 T2 

Since all parameters on right hand side are known, a can be estimated.

…(1.37)

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29

Method II.

In this alternative method, Sabine used an organ pipe of frequency 512 Hz and measured the reverberation time of the hall with a large amount of area of the material whose absorption coefficient is to be measured being kept inside along with the cushion, carpet etc. After measuring the reverberation time, the material whose absorption coefficient is to be found is removed from the hall. This will naturally result in increase in the reverberation time of the hall. Now, by continuously adjusting (increasing) the area of open windows present in the hall, the reverberation time of the hall is found. This procedure is followed until the reverberation time of the hall becomes equal to the reverberation time of the hall when the material/surface whose absorption coefficient is to be measured was inside the hall. At this stage, the increase in area of the open window of the hall that was necessary to make the reverberation time values equal is measured. Then absorption coefficient of the material (a) is calculated using the formula a=

Increase in area of open window …(1.38) Area of material whose absorption coefficient is to be measured

Method III.

The third method discussed here is for estimating the average absorption coefficient of a hall. This method is based on statistical calculations made by Jaeger to derive an expression for the reverberation time of a hall. Let V and S represent the volume of the hall and the total area of all absorbing surfaces/materials in the hall, respectively. Let I0 be the intensity of sound just when it is stopped and v, be the speed of sound respectively. Let It be the intensity of sound t second after the sound source is stopped. If ∝ is the average absorption coefficient of the hall. Jalger has proved that It = I 0

α Svt e 4V

…(1.39)

Now let us assume that there are two sources of sound in the hall. Let I1 and I2 be the intensity of sound produced by these two sources respectively just when they are stopped. Let P1 and P2 be the effective pressures exerted in the medium by the sound waves from the two sources respectively, just when they are stopped. Now it is clear that when the intensity of two sounds, namely I1 and I2 fall to one-millionth of their values, they will become inaudible (i.e., reach threshold of audibility) and the time taken by these sound waves to reach this level is called reverberation time. Hence if T1 and T2 are the reverberation time of the sound waves from the two sources and if the intensity of sound just when it becomes inaudible is represented by Ith (where Ith refers to threshold intensity), we may write I th = I1e

–

α SvT1 4V

…(1.40)

–

α SvT2 4V

…(1.41)

and I th = I 2 e

30


for the sound from the two sources. Equating the right hand sides of Eqns. (1.40) and (1.41),

I1e

–

α SvT1 4V

= I2e –

–

α SvT2 4V

α SvT2

e 4V I1 = α SvT1 I2 – e 4V

Hence

α Sv

– I1 = e 4V I2

or

(T1 −T2 )

Taking natural logarithm on both sides α Sv I  (T1 – T2 ) loge  1  = 4V  I2 

\

α =

I  4V × loge  1  Sv(T1 – T2 )  I2 

i.e.,

α =

I  4V × 2.303 × log10  1  Sv(T1 – T2 )  I2 

…(1.42)

But, we know that, from Eqn. (1.11) (sound pressure level) 2

I1  P1  =   …(1.43) I2  P2  where P1 and P2 are the effective pressures of the plane progressive sound waves from source 1 and source 2 respectively. Substituting Eqn. (1.43) in Eqn. (1.42), we get P  4V × 2.303 × log10  1  α = Sv(T1 – T2 )  P2  α =

2

P  8V × 2.303 × log10  1  Sv(T1 – T2 )  P2 

…(1.44)

Since all quantities on the right hand side of Eqn. (1.44) can be measured, the average absorption coefficient of hall can be measured.

PROBLEMS AND SOLUTIONS IN ACOUSTICS Example 1.5 Calculate the reverberation times of a hall of 2400 m3 having a seating capacity of 600, (a) when the hall is empty, and (b) when the hall is full with with audience from the following data.

I


Surface/description

31

Area or number

Absorption coefficient (O.W.U.)

Plastered ceiling

500 m2

0.02

Plastered wall

600 m2

0.03

Wooden floor

500 m2

0.06

Wooden doors

20 m2

0.06

Cushioned seats

400

0.1 per chair

Cane seats

200

0.01 per chair

Audience

—

0.45

Solution (a) When the hall is empty T=

0.16V Σα i dsi

=

0.16 × 2400 500 × .02 + 600 × .03 + 500 × .06 + 20 × .06 + 400 × .1 + 200 × .01

=

384 10 + 18 + 30 + 1.2 + 40 + 2

384 101.2 = 3.79 s \ When the hall is empty, T = 3.79 s (b) When the hall is full with audience

=

T=

0.16V ∑ α i dsi

=

0.16 × 2400 500 × 02 + 600 × .03 + 500 × .06 + 20 × .06 + 400 × .45 + 200 × .45

=

384 10 + 18 + 30 + 1.2 + 180 + 90

384 329.2 = 1.17 s \ When the hall is full, T = 1.17 s

=

Example 1.6 Calculate the reverberation time of a hall of 1800 m3 having a seating capacity for 160 persons (i) when the hall is empty, (ii) when the hall is full with audience, and (iii) when the cane chairs are alone occupied by audience, using the following data.

32


Surface or description

Area or number

Absorption coefficient (O.W.U)

150 m2

0.05

Plastered walls

100

m2

0.025

Plastered ceiling

2.20 m2

0.03

Wooden floor

Wooden door

50

m2

0.04

Cushioned chairs

50

0.08 per chair

Cane chairs

80

0.01 per chair

Audience

—

0.4

Solution (i) When the hall is empty

0.16V = Σα idsi 150 × .05 + 100 × .025 + 220 × .03 + 50 × .04 + 80 × .08 + 80 × .01 0.16 × 1800 = 150 × .05 + 100 × .025 + 220 × .03 + 50 × .04 + 80 × .08 + 80 × .01

T =

=

288 7.5 + 2.5 + 6.6 + 2.0 + 6.4 + 0.8

288 25.8 = 11.163 s \ When the hall is empty, T = 11.136 s (ii) When the hall is full with audience

=

T=

0.16V Σα i dsi

0.16 × 1800 150 × .05 + 100 × .025 + 220 × .03 + 50 × .04 + 80 × .4 + 80 × .4 288 = 7.5 + 2.5 + 6.6 + 2.0 + 32 + 32

=

288 82.6 = 3.487 s \ When the hall is full with audience T = 3.487 s (iii) When the cane chairs are alone occupied

=

T= =

0.16 V Σα i dsi

0.16 × 1800 150 × .05 + 100 × .025 + 220 × .03 + 50 × .04 + 80 × .08 + 80 × .4

I


33

=

288 7.5 + 2.5 + 6.6 + 2.0 + 6.4 + 32

288 57 = 5.053 s \ When cane chairs are alone occupied, T = 5.053 s

=

Example 1.7 The reverberation time of a theatre has to be designed to be 2.1 s. If the length, breadth and height of the theatre are 25 m, 20 m and 8 m respectively, how much should be the total absorption inside the theatre? Solution Volume of theatre, V = 2.5 × 20 × 8 = 4000 m3 Reverberation time, T = 2.1 s Sabine’s formula for reverberation time is T=

0.16V Σα i dsi

\ Total absorption inside theatre

Σα i dsi =

=

0.16V T 0.16 × 4000 2.1

640 2.1 = 304.8 O.W.U. m2 = 304.8 O.W.U. m2

= \ Total absorption inside the theatre Example 1.8

The reverberation time of a hall of volume 3000 m3 is 4.6 s. Given a material of absorption coefficient 0.69 O.W.U., how will you reduce the reverberation time of the hall to 2.7s? Solution Let T1 be the initial reverberation time of the hall. Then T 1 = 4.6 =

0.16 V Σα i dsi

…(1.45)

If we increase the absorption inside the hall, the reverberation time can be brought down. Hence let an area S of given material with absorption coefficient a = 0.69 O.W.U. be introduced in the hall so that the final reverberation time T2 is 2.7 s. \

T 2 = 2.7 =

0.16 V Σα i dsi + αs

…(1.46)

34


From Eqns. (1.46) and (1.45), we have

1 1 αs – = T2 T1 0.16 V \

S= =

0.16V  1 1   –  α  T2 T1  0.16 × 3000  1 1  –   0.69  2.7 4.6 

0.16 × 3000 × .153 .69 \ S = 106.435 m3 By introducing 106.4 m2 area of given material inside the hall the reverberation time can be reduced to 2.7 s.

=

Example 1.9 The reverberation time of an auditorium of volume 4300 m3 is 1.9 s. It is required that the reverberation time should be increased to 2.34 s. If the absorption coefficient of one of the absorbing materials inside the hall is 0.48 O.W.U., how will you use that material to achieve a reverberation time of 2.34 s? Solution Initial reverberation time T 1= 1.9 =

0.16 V Σα i dsi

…(1.47)

We know that if the total absorption in the hall is reduced, the reverberation time can be increased. Hence let us assume that we remove an area S of the absorbing material of absorption coefficient µ = 0.4 s O.W.U. to get a reverberation time of 2.34 s. \ Final reverberation time, T2 = 2.34 =

0.16 V Σα i dsi – α s

…(1.48)

From Eqns. (1.47) and (1.48), we have

αs 1 1 – = 0.16 V T1 T2 \

S= =

0.16V  1 1   –  α  T1 T2  0.16 × 4300  1 1  –   0.48  1.9 2.34 

0.16 × 4300 × 0.099 0.48 = 141.9 m2 By removing 141.9 m2 area of an absorbing material of absorption coefficient 0.48, we can increase the reverberation time from 1.9 s to 2.34 s.

=

I


35

Example 1.10 Calculate the intensity of sound at a given place if the frequency of sound is 1000 Hz, amplitude of vibrating particles is 10–11m. Given that the density of air and velocity of sound in air at room temperature are 1.3 kg/m3 and 350 m/s respectively. Solution Intensity of sound,

I = 2π 2 a 2 n 2 eV = 2 × π 2 × (10 –11) 2 × (1000) 2 × (1.3) × 350 = 8.98 × 10 –13 W/m 2

Ans. Intensity of sound

= 8.98 × 10 –13 W/m2

Example 1.11 What is the difference in intensity levels of two sounds whose intensities are in the ratio 1:100? Solution Let I1 and I2 be the intensities of two sources. Then I 2 = 100 I1. Let IL1 and IL2 be the intensity levels due to the two sources. I  I IL2 – IL1 = 10 log10  2  – 10 log10  1 I    0  I0 where I0 = reference intensity = 10–12 W/m2 From Eqn. (1.49), we have

\

   

  100 I1   I   – log10  1  IL2 – IL1 = 10log10     I0   I 0    100 I1 I 0 × = 10log10  I1   I0 = 10 log10 100 = 10 log10102 = 20 log1010 = 20 dB

  

Ans. Difference in intensity level = 20 dB Example 1.12 Calculate the loudness of a sound of intensity equal to 10 times the reference intensity. Solution Intensity of given source of sound, I = 10 I0, where I0 is the reference intensity.

…(1.49)

36


 I  \ Loudness (or) intensity level = 10 log10    I0   10 I 0 = 10 log10   I0 = 10 log10 10 = 10 dB

Ans. Loudness of given source

   

= 10 dB

Example 1.13 Calculate the intensity level produced by a sound intensity of 100 I0, where I0 is the reference intensity. Solution Given intensity of sound \

I = 100 I0

  Intensity level = 10 log10  I  I   0  100 I 0 = 10 log10   I0

= = = =

   

10 log10 100 10 log10 102 20 log10 10 20 dB

Ans.

Intensity level produced by   = 20 dB a source of intensity100 I 0  Example 1.14 When a sound source A is ‘ON’, the intensity level at a given place is found to be 70 dB. When another source B alone is ‘ON’, the intensity level at the same place is 20 dB. Calculate the intensity level at the place when both the sources are ‘ON’. Solution Let I1 be the intensity of sound waves coming from source A. Let ILA be the intensity level due to source A. Then

I  IL A = 10 log10  1   I0  I  70 = 10 log10  1  I   0

I


37

I  7 = log10  1  I   0 I1 antilog 7 = I0 \ I 1 = (antilog 7) × I0 = 107 × I0 = 107 × 10–12 I 1 = 10–5 W/m2 Let I 2 be the intensity of sound waves from source B. Let IL B be the intensity level due to source B.

Then

I IL B = 10 log10  2 I  0

   

I 20 = 10 log10  2  I0

   

 I2  2 = log10    I0  I antilog 2 = 2 I0 \ I 2 = (antilog 2) × I0 = 10 2I 0 = 102 × 10–12 \ I 2 = 10 –10 W/m2 Hence when both the sources are ‘ON’, the resultant intensity I = I1 + I2. \ I = 10–5 + 10–10 = 10 –10(10 5+1) I = 10 –10(100001) W/m2

\ Resultant intensity level at   I   the given place when both  IL = 10 log10    I0   the sources are ‘ ON’  10 –10 × 100001  = 10 log10   10 –12   = 10 log10 (102 × 100001) = 10 (2 log10 10 + log10 100001) = 10 (2 + 5 ) = 10 × 7 IL = 70 dB Ans. Resultant intensity level

= 70 dB

38


Example 1.15 What would be the resultant intensity and resultant intensity level when two sounds each of 40 dB combine? Solution Given IL 1 = 40 dB and IL2 = 40 dB. Let I be the intensity of sound that can produce an intensity level or loudness 40 dB. Then  I  40 = 10 log10    I0   I  4 = log10    I0  I antilog 4 = I0

\

I = (antilog 4 ) × I0 = 104 × 10–12 = 10–8 W/m2

When two sounds of 40 dB each  \ combine, the resultant intensity  IR = I + I  = 10–8 + 10–8 or I R = 2 × 10–8 W/m2 Now let us find the resultant intensity level when two sources of 40 dB each combine.

We know that the resultant intensity  –8 2 when two sounds each of 40 dB combine IR = 2 × 10 W/m I \ Resultant intensity level when two sources  IL = 10 log10  R each of 40 dB combine is given by   I0

   

 2 × 10 –8   = 10 log10   10 –12   

= = = = = =

10 log10 (2 × 104) 10 [ log10 2 + log10 104] 10 [ 0.3010 + 4 log10 10] 10 (0.3010 + 4] 10 × 4.3010 43.01 dB

I


39

Ans. (i)

When two sounds of 40 dB each  I = 2 × 10–8 W/m2 combine, the resultant intensity  R

(ii)

When two sounds each of 40 dB   combine, the resultant intensity  IL = 43.01 dB.  level

Example 1.16 Certain sound produces a loudness of 30 dB at a given place. What should happen to the intensity of sound in order to double the loudness at that place? Solution Let I1 be the intensity of sound corresponding to 30 dB intensity level. Then I 30 = 10 log10  1  I0 I  3 = log10  1   I0  I antilog 3 = 1 I0

\

   

I 1 = (antilog 3 ) × I0 = 103 × 10–12 I 1 = 10–9 W/m2

Given that final intensity level = 2 × 30 = 60 dB. Let I2 be the final intensity of sound for 60 dB loudness. Then

\

I  60 = 10 log10  2   I0  I 2 = (antilog 6) × I0

= 106 × 10–12

\

I 2 = 10–6 W/m2 I2 10 –6 = 103 = 1000 = –9 I1 10

Note : From Example 1.14, we understand that when two intensity levels 70 dB and 20 dB act simultaneously, the resultant intensity level is 70 dB only. In Example 1.15, we have seen that when two sound sources each of 40 dB combine, the resultant intensity level is 43 dB only and it is not 40 + 40 = 80 dB. From Examples 1.14 and 1.15, we conclude that we should not add numerically two intensity levels to get the

40


\

I 2 = 1000 I1

Ans. When the intensity of sound is increased to 1000 times its initial intensity, the intensity level would change from 30 dB to 60 dB. Example 1.17 Calculate the ratio of final intensity to the initial intensity of sound when the intensity level of sound changes from 50 dB to 100 dB. Solution Intial intensity level IL i = 50 dB Initial intensity of sound = Ii Then

\ or

I  50 = 10 log10  i   I0  I  5 = log10  i   I0  Ii antilog 5 = I0 I i = (antilog 5) × I0 = 105 × 10–12

I i = 10–7 W/m2

Given that final intensity level IL f = 100 dB Let final intensity of sound = I f

\

If 100 = 10 log10   Ii I f = (antilog 10)

or

I f = 10–2 W/m2

Then

If

    × I0

10 –2

= 10 5 –7 Ii 10 Ans. The ratio of final intensity of sound to the initial intensity of sound is 105 when the intensity level is doubled from 50 dB to 100 dB. \

=

Example 1.18 The intensity of sound at a given place due to a source of sound is 3.5 × 10–2 W/m2. Estimate the intensity level at that place in decibel. Solution Given that \ Intensity level

I = 3.5 × 10–2 W/m2  I  IL = 10 log10    I0 

I


41

 3.5 × 10 –2   = 10 log10   10 –12   

= 10 log10 (3.5 × 1010) = 10 (log10 3.5 + log10 1010) = 10 (0.54 + 10 log10 10) = 10 (0.54 + 10) = 10 (10.54) = 105.4 dB Ans.

Intensity level of sound for   = 105.4 dB an intensity of 3.5 × 10 –2 W/m2  Example 1.19 By how much would the intensity level at a given place change when the intensity of sound produced by a source at that place is doubled? Solution Let initial intensity level of sound be IL1. Let I1 be the initial intensity of sound. \

I  IL 1 = log 10  1   I0 

It is given that final intensity of sound I2 = 2 I1. Let final intensity level of sound be IL2. Then Final intensity level

I  IL 2 = 10 log10  2   I0   2I  = 10 log10  1   I0   I = 10 log10  2 × 1 I0 

   

I = 10 log10 2 + 10 log10  1  I0 = 10 × 0.3010 + IL1

\

   

IL 2 = 3.01 + IL1 dB

\ Change in intensity level = IL2 – IL1 = (3.01 + IL1) – IL1 = 3.01 dB Ans. The intensity level is increased approximately by 3 dB whenever the intensity of sound is doubled.

42


Example 1.20 Calculate the intensity of sound of frequency 1000 Hz. Corresponding to the threshold of feeling or pain, namely, 120 dB. Solution Given that the threshold of feeling (or) intensity level of sound that produces pain on the ear = 120 dB. Let I be intensity of sound which produces an intensity level of 120 dB. Then  I  IL = 10 log10   becomes I   0  I  120 = 10 log10    I0   I 12 = log10   I0

   

I antilog 12 = I 0 \

I = (antilog 12) × I0 = 1012 × 10–12 = 100 I = 1 W/m2

Ans. Intensity of sound corresponding to threshold of feeling = 1 W/m2.

OBJECTIVE QUESTIONS IN ACOUSTICS (OA) OA 1. The (a) (b) (c) (d)

intensity of sound: and loudness have same two units cannot be measured depends on loudness is measured in W/m2

OA 2. Mosquitoes produce: (a) sounds of low intensity (b) sounds of high pitch (c) ultrasonics only (d) infrasonics only OA 3. Pitch of a sound: (a) is an objective quantity (b) is a physical quantity

I


(c) will be same irrespective of listener (d) cannot be measured OA 4. Loudness of sound: (a) is directly proportional to intensity (b) and intensity of sound are same (c) increases with increase in intensity but not proportionally (d) has a unit of W/m2 OA 5. Intensity of sound: (a) is directly proportional to square of frequency (b) is inversely proportional to density of medium (c) does not depend on speed of sound in the medium (d) is inversely proportional to frequency of sound OA 6. Loudness of sound: (a) is proportional to square of intensity (b) is proportional to logarithm of frequency (c) is inversely proportional to intensity (d) is proportional to the logarithm of intensity OA 7. The (a) (b) (c) (d)

intensity of sound corresponding to threshold of audibility is: 10–12 W/cm2 for 1000 Hz frequency 10–12 W/m2 for 1000 Hz frequency in air 10–12 W/m2 for 512 Hz frequency in air 10–16 W/m2 for 1000 Hz frequency

OA 8. The (a) (b) (c) (d)

intensity of sound of 1000 Hz frequency corresponding to threshold of feeling is: 1012 W/m2 0 W/m2 1 W/m2 10 –12 W/m2

OA 9. The (a) (b) (c) (d)

reference intensity value used for calculating intensity level is: 10 –16 W/m2 10 –12 N/m2 10–12 W/cm2 10 –12 W/m2

OA 10. The (a) (b) (c) (d)

reference intensity of sound produces: one decibel one watt zero decibel 120 dB

43

44


OA 11. Different sounds of equal intensity and of different frequencies will: (a) produce equal loudness (b) not produce equal loudness (c) will have same pitch (d) will have deafening effect on ear OA 12. A sound of frequency 30 Hz is audible: (a) if its intensity is 10–12 W/m2 (b) if its intensity is 10–8 W/cm2 (c) if its intensity is 10–12 W/cm2 (d) if its intensity is 10–6 W/m2 OA 13. The (a) (b) (c) (d)

intensity of sound for a plane progressive wave is directly proportional to: the effective pressure exerted by sound wave the square of effective pressure density of medium the speed of sound in the medium

OA 14. If P and P0 represent the effective pressure and reference pressure of sound waves, then sound pressure level is given by: P (a) 10 log10   decibel  P0  P (b) 20 log10   bel  P0  P (c) 10 log10   P0

(d)

P 20 log10   P0

  bel     decibel  

OA 15. The value of effective pressure corresponding to threshold of audibility of a sound of frequency 1000 Hz in air is: (a) 10–12 N/m2 (b) 10–5 W/m2 (c) 2 × 10–5 N/m2 (d) 2 × 10–12 W/m2 OA 16. When same music is played in veena and violin, we shall hear sounds of: (a) same intensity (b) same quality (c) different quality (d) same quality and same intensity

I


OA 17. The (a) (b) (c) (d)

unit of reverberation time is: bel decibel second W/m 2

OA 18. Unit (a) (b) (c) (d)

of absorption coefficient is: O.W.U. joule watt W/m 2

45

OA 19. The frequency of muscial echo produced by flight of steps of width d for normal incident sound waves (v is the speed of sound) is: 2d (a) v v (b) d 2v (c) d (d)

v 2d

OA 20. Whenever the intensity of sound is doubled: (a) the loudness level is also doubled (b) the intensity level is halved (c) the intensity level increases by 100 times (d) the intensity level increases by 3 dB

KEY FOR OBJECTIVE QUESTIONS IN ACOUSTICS OA 1. d

OA 2. b

OA 3. d

OA 4. c

OA 5. a

OA 6. d

OA 7. b

OA 8. c

OA 9. d

OA 10. c

OA 11. b

OA 12. d

OA 13. b

OA 14. d

OA 15. c

OA 16. c

OA 17. c

OA 18. a

OA 19. d

OA 20. d

SHORT QUESTIONS IN ACOUSTICS (SA) SA 1. How is sound classified? SA 2. What is pitch?

46


SA 3. SA 4. SA 5. SA 6. SA 7. SA 8. SA 9. SA 10. SA 11. SA 12. SA 13. SA 14. SA 15. SA 16. SA 17. SA 18. SA 19. SA 20. SA 21. SA 22. SA 23.

What happens to pitch of a sound when frequency is increased? Does loudness of sound change when the intensity of sound increases? Define intensity of sound. What is its unit? Does area of a sounding body influence the loudness? Justify your answer. Explain inverse square law in sound. Explain Weber-Fechner law. What is meant by threshold of audibility? Explain threshold of feeling. What is the value of threshold intensity for a sound of frequency 1000 Hz? How many decibel of sound approximately causes pain to human beings? What is meant by quality of sound? Explain phon. Explain sone. Discuss “sound pressure leve”. What is the value of effective pressure corresponding to threshold of audibility? Define reverberation. Define reverberation time. Define reverberation time in terms of decibel. State Sabine’s formula for reverberation time and explain the terms involved. Define absorption coefficient. Explain the meaning of one decibel.

REVIEW QUESTIONS IN ACOUSTICS (RA) RA 1. RA 2. RA 3. RA 4. RA 5. RA 6. RA 7.

Explain characteristics of musical sound. Discuss Weber-Fechner law and parameters which affect loudness of sound. Define absorption coefficient and discuss any one method of measuring the same. Explain different methods of measuring absorption coefficient. Derive Sabine’s formula for reverberation time. Discuss the growth and decay of energy density of sound. Explain and discuss the following: (i) Loudness (ii) Focussing due to walls (iii) Echo (iv) Echelon effect (v) Resonance (vi) Noise

I


47

RA 8. Write notes on acoustics of buildings. RA 9. Discuss any four factors affecting acoustics of buildings and their remedies. RA 10. Define reverberation time and discuss the method of measuring the same.

EXERCISE ON ACOUSTICS (EA) EA 1. Calculate the intensity of sound waves of amplitude 6 × 10–4 W/m2. (Density of air is 1.3 kg/m3 and speed of sound is 340 m/s). EA 2. Estimate the intensity level of sound whose intensity is 10–8 W/m2. EA 3. Calculate the ratio of final intensity to initial intensity of sound when the intensity level of sound changes from 60 dB to 120 dB. EA 4. Calculate the ratio of final intensity to initial intensity of sound when the intensity level of sound changes from 20 dB to 40 dB. EA 5. Find the difference in intensity levels of sound when the intensity of sound is tripled. EA 6. Prove that 26% increase in intensity of sound can cause a change in intensity level by 1 dB. EA 7. Calculate the total absorption in a cinema hall of volume 3000 m3 and 2.2 s reverberation time. EA 8. The reverberation time of a hall of volume 1000 m3 is 3.2 s. How will you reduce the reverberation time to 2.3 s using a material of absorption coefficient 0.39? EA 9. The reverberation time of a hall of volume 2500 m3 is 1.9 s. How will you increase it to 2.2 s using a material of absorption coefficient 0.45 O.W.U.? EA 10. Calculate the reverberation time of a hall of volume 4200 m3 with a seating capacity of 200 when (i) the hall is empty, (ii) the hall is full with audience, and (iii) 50% of caned chairs and 50% of cushioned chairs are alone occupied, using the data given below: Item/ Description

Area or Number

Absorption Coefficient (O.W.U.)

Plastered walls

300 m2

0.02

150

m2

0.03

80

m2

0.05

Cushioned chairs

100

0.8

Cane chairs

100

0.3

Plastered ceilings Wooden doors

Audience

—

0.45

48


B. STRUCTURE OF SOLIDS 1.7 INTRODUCTION All elements and their chemical compounds take one of the three states viz. solids, liquids and gases. These materials are composed of atoms or ions or molecules. However, in solids and liquids the spacing between neighbouring atoms is of the order a few Angstroms with about 1028 atoms/m3 while that in gases at room temperature, under one atmospheric pressure, the average spacing is of 30 Angstrom with about 1025 molecules/m3. Remember, not all solids are crystals. Some solids have no periodic structure at all and are called amorphous or glassy. These include glass, fused silica, rubber, polymers etc. Some solids which have three-dimensional periodic arrangement of their constituents (namely atoms or molecules or ions) are called crystals. Aluminium, copper, zinc etc. are some examples of crystalline solids.

1.8 SPACE LATTICE AND TRANSLATION VECTORS Space Lattice A lattice is a regular periodic arrangement of points in space that has a net-like structure. This arrangement of points is such that environment around any one point is identical with the environment around any other point. Figure 1.10 shows such an array of points in two dimensions. If points are arranged at regular intervals in a row or line, then we get a row lattice. In Fig. 1.10, OA and OB are row lattices with an arbitrary point ‘O’ being chosen as origin. Along OA the points are repeated at a regular interval “a”. This repetitive distance “a” along the direction OA is represented by →

“translation vector” or “lattice vector” or “primitive”, namely, a . Along OB, the repetitive distance is →

‘b’ and the primitive of lattice vector is represented by b . B

b O

a

A

b a Fig. 1.10 Two dimensional array of points

It may be noted that the angle between OA and OB need not necessarily be 90o. If we extend the array of points to three dimensions, then the array of points is called the space lattice.

Translation Vectors If we choose any lattice point as origin (say ‘O’ in Fig. 1.10), then the position vector of any other lattice point is represented by

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49

H H H ...(1.50) T = n1a + n2 b where n1 and n2 are integers representing the number of intervals taken along OA and OB respectively. →

→

a and b are the intervals or “translation vectors” or “lattice vectors” or “primitives”, along OA and OB respectively [Fig. 1.10]. For a three-dimensional lattice, the arrangement of points can similarly be represented by translational (or lattice) vector. H H H H ...(1.51) T = n1a + n2 b + n3 c →

→

→

where a, b and c are fundamental translation vectors (or basis vectors) or called primitives along the three directions in which the points are repeated.

1.9 UNIT CELL AND UNIT CELL PARAMETERS → → →

The parallelopiped formed by using the basis vectors (or primitives) a , b , c as concurrent edges is called the unit cell of space lattice as shown in Fig. 1.11. It is the smallest closed figure, formed by the least number of lattice points which is fully representative of the lattice and which when repeated over → → →

and over again in different directions a , b , c forms the actual lattice itself. The angles included → →

→

→

→

→

between primitives a and c , c and a, a and b are denoted by a, b and g respectively. The parameters a, b, c, a, b, g which define the size and shape of a unit cell are called unit cell parameters. Z

(0,0,c) c α

β 0

X

a (a,0,0)

γ

Y b

(0,b,0)

Fig. 1.11 Three-dimensional unit cell

1.10 BASIS AND CRYSTAL STRUCTURE In most elements, there is a single atom (or ion) at each lattice point. In others (compounds), there is a group of two or more atoms (or ions) at each lattice point. This group of atoms is called a basis. It is an assembly of one or more atoms which when positioned at each lattice point gives rise to crystal structure. Every basis is identical in composition, arrangement and orientation. In crystals like aluminium

50


and barium the basis is a single atom. In NaCl, KCl, ZnS etc., the basis is diatomic whereas in crystals like BaF2 and CaF2, the basis is triatomic. Basis is repeated in space lattice to form the crystal. When a basis is added to each lattice point in space, crystal structure is generated. The logical relation is Lattice + Basis ® Crystal structure Formation of the structure of crystal by positioning a basis at every point of a lattice is shown in the Fig. 1.12.

Fig 1.12 (a) Lattice (b) Basis containing two different ions ( c) Crystal structure

1.11 PRIMITIVE AND MULTIPLE CELLS A unit cell which has only one lattice point is called a primitive cell. For example let us consider a simple cubic unit cell with one lattice point at each of the corners. Then it appears as though the cube contains eight lattice points. But each corner of the cubic unit cell is shared by eight unit cells. Hence lattice point at each corner contributes only one-eighth of a lattice point to the unit cell. Since there are 1 eight such corners, the effective number of lattice points in the unit cell is 8 × = 1. Hence a simple 8 cubic unit cell is called a primitive cell.

On the other hand, let us consider a cubic unit cell with one lattice point each at each one of the corners and one lattice point at its body center. Now as already seen the corner lattice points contribute one lattice point to the unit cell. The lattice point at the body center is wholly owned by the given unit cell and is not shared by any other unit cell. Hence the effective number of lattice points in the unit cell 1 + 1 = 1 + 1 = 2. Such a unit cell is called a multiple cell since it has more than one lattice point in 8 it. This unit cell may also be called a doubly primitive cell. A knowledge of effective number of lattice points of a unit cell helps us to understand and estimate the density of packing of a unit cell. It is important to understand that all primitive cells are unit cells. But all unit cells are not primitive cells. Out of the 14 Bravais lattices, there are only 7 primitive unit cells and the remaining are multiple cells (Refer Table 1.2).

is 8 ×

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51

Table 1.2 Crystal systems and Bravais lattices Relation between Sl. No

System Axes

Bravais lattices

Angles

Lattice Number of symbol lattices in the system

Cubic

a=b=c

a = b = g = 90°

Simple, body centred, face centred

P I F

3

1.

2.

Tetragonal

a=b¹c

a = b = g = 90°

Simple, body centred

P I

2

3.

Rhombohedral (Trigonal)

a=b=c

a = b = g ¹ 90°

Simple

P

1

4.

Hexagonal

a=b¹c

a = b = 90° and g= 120°

Simple

P

1

P C I F

4

5.

Orthorhombic (Orthogonal)

a¹b¹c

a = b = g = 90°

Simple, base centred, body centred, face centred

6.

Monoclinic

a¹b¹c

a = b = 90° ¹ g

Simple, base centred

P C

2

7.

Triclinic

a¹b¹c

a ¹ b ¹ g ¹ 90°

Simple

P

1

Total

7

Total

14

1.12 BRAVAIS LATTICES A study of the parameters of the unit cell reveals that crystals fall into any one of the seven systems on the basis of the shape of the unit cells. The seven crystal systems and relation between unit cell parameters of these ststems are shown in Table 1.2. Bravais, in 1948, showed that there are fourteen different types of lattices under the seven crystal systems. In general, a unit cell which has lattice points only at the corners is called a simple cell or a primitive cell and it is represented by the symbol P. If a unit cell has one lattice point each at the center of all faces, then it is called face centred cell and is represented by the symbol F. If a unit cell has one lattice point at its body center in addition to its lattice points at the corners, it is called body centred cell and is represented by I. All the seven crystal systems, put together, have only 14 types of cells. These 14 unit cells are called Bravais lattices. Thus we have 14 Bravais lattices. These Bravais lattices have established that there are only 14 distinguishable ways of arranging points in three-dimensional space. For a cubic system, we have three types of lattices. The first one is simple cubic or P-type lattice with lattice points only at the eight corners of the unit cell. The second one is body centred cubic with one more additional lattice point situated at the body center, well within the unit cell. This is called I type lattice. The last one is face centred cubic or F lattice. Here, in addition to the eight lattice points, one at each corner, there is an extra lattice point at the centre of each face.

52


1.13 COORDINATION NUMBER With reference to a lattice point in a unit cell, the number of nearest neighbors i.e., the number of lattice points at the closest distance is called the coordination number. This also helps one to understand how densely a unit cell is packed. If we consider a simple cubic unit cell of edge ‘a’, then with reference to any one lattice point at the corner of the cell, there can be six lattice points each at a distance ‘a’ from it. Then the coordination number or number of nearest neighbors of a simple cubic unit cell is 6. For a face centered cubic unit cell of edge ‘a’, each lattice point at the corner has 12 nearest a neighbors, the nearest neighbor distance being along the face diagonal (4 lattice points each in ab 2 plane, ac plane and bc plane make 12 neighbors). Hence coordination number for fcc is 12. For a body centred cubic unit cell of edge ‘a’, each corner lattice point has one lattice point at a a 3 distance of along the body diagonal. Since each corner lattice point is also in contact with 8 such 2 a 3 each from the corner along the body unit cells, there are 8 lattice points at the distance of 2 diagonal. Hence coordination number for bcc is 8. If we consider an hcp unit cell, the central lattice point at the top plane or at the bottom plane is in contact with 6 lattice points each at a distance of ‘a’ in its own plane. Further it will be in contact with 3 lattice points above it and below it, each at a distance of ‘a’. Hence coordination number for hcp is 12.

1.14 ATOMIC PACKING FACTOR (APF) OR PACKING FRACTION (PF) OR RELATIVE DENSITY OF PACKING (RDP) The unit cells of crystals are not always fully occupied by the atoms or ions. The atoms and ions or basis which occupy lattice points leave lot of empty space in the unit cell. Hence it is also important to know the density of packing. This is done by defining and measuring Atomic Packing Factor (APF) or Packing Fraction (PF) or Relative Density of Packing (RDP). Atomic Packing Factor (APF) is defined as the ratio of volume of atoms in the unit cell to the volume of the unit cell. Atomic Packing Factor (APF) or Packing Fraction (PF) or Relative Density of Packing (RDP) =

Volume of atoms in the unit cell volume of unit cell

number of atoms in the unit cell × volume of one atom volume of unit cell v = ...(1.52) V From the above it is also clear that atomic packing factor or packing fraction may also be defined as the volume occupied by atoms in unit volume of the unit cell. We shall discuses the atomic packing fraction of the cubic and hexagonal close packed structures as they find a number of applications in practice. =

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53

1.14.1 Atomic Packing Factor (APF) or Packing Fraction (PF) of Cubic Structures and Hexagonal Close Packed Structure Cubic crystal is the simplest of all and is found to have three structures, and are normally described as simple (i.e., primitive), body centred and face centred cubic. 1. Simple Cubic (SC) Structure (Primitive)

This structure is shown in Fig. 1.13. Eight atoms are arranged in a regular cube, with one atom at each corner. It is a cubic unit cell of size ‘a’. The arrangement of atoms is repeated in all x, y and z directions. It may be understood that each corner atom is shared by 8 unit cells i.e., each one of the 1 i.e., one-eighth of itself to the unit cell. Hence the atoms at the eight corners contributes only 8 1 number of atoms in the unit cell is 8 × = 1. Also a = 2r since the atoms touch along the edge as shown 8 in Fig. 1.13. We know that Atomic Packing Factor (APF) is the volume occupied by atoms in unit volume of unit cell or it is the ratio of the volume of atoms in a unit cell to the volume of unit cell. This is also called Relative Density of Packing (RDP).

a =2r

Fig. 1.13 Simple cubic structure

Thus Atomic Packing Factor (APF) or Packing Fraction (PF) or Relative Density of Packing (RDP) =

Volume of atoms in the unit cell volume of unit cell

=

number of atoms in the unit cell × volume of one atom volume of unit cell

v ...(1.52) V where v is the volume of atoms in the unit cell and V is the volume of unit cell. In this simple cubic (sc) structure, the number of atoms in a unit cell is 1. If r is the radius of the atom, a = 2r and volume of unit cell, V = a3.

=

 4 3 1 × 3 π r  v  ∴ APF = = V a3

54


=

4π r 3 3 × (2r ) 3

π = 0.52 6 APF = 52%

=

...(1.53)

Polonium is the only element which has simple cubic structure. 2. Body-centred Cubic (BCC) Structure

There is one atom at each corner of the cube and one at the centre of the unit cell (entirely within the unit cell). Iron (at room temperature), chromium, tungsten etc. have bcc structure. See Fig. 1.14.

r F

F

a

A

C

r

2r

4r

a

C

a

Fig. 1.14 Body centred cubic structure

1 ×8 +1= 2 8 CF = 4r (CF) 2 = (CA)2 + (AF)2 (4r)2 = (CA)2 + a2 = 2a2 + a2 = 3a2 4r ∴ 4r = 3a; a = 3

Number of atoms in a unit cell = Body diagonal, From DCAF,

v = 2×

\

3 4 3 8πr πr = 3 3 v APF = V

=

and V = a 3 =

64 r 3 3 3

8× πr3 × 3 3 3 × 64 r 3

π 3 = 0.68 8 APF = 68%

=

...(1.54)

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55

3. Face-centred Cubic (FCC) Structure

There is one atom at each corner of the cube and one atom at the centre of each face. Copper, gold, nickel, lead etc. are of fcc structure. The atom at the center of each face is shared by two unit cells, or we can say that each face of the fcc unit cell contains 1/2 atom. Refer Fig. 1.15.

4r

a

a Fig. 1.15 Face centred cubic structure

The number of atoms in FCC unit cell is 1 1 8× + 6× = 4 8 2 Also from the figure, (4r)2 = a2 + a2; 16r2 = 2a2 or a=

4r 2

\

v APF = V

4 4× πr3 3 = a3 = =

16π r 3 × 2 2 3 × 64 r 3

π 3 2

= 0.74

APF = 74%

...(1.55)

4. Hexagonal Close Packed (hcp) Structure

Elements do not crystallize in the simple hexagonal form but many of them crystallize in the hexagonal close-packed (hcp) structure which is shown in Fig. 1.16. Hexagonal close packed structure is of hexagonal shape of side ‘a’ and height ‘c’. Each corner atom is shared by six other unit cells or each corner contributes 1/6 atom to the units cell. Hence the number of atoms in the upper and lower planes

56


1 1 × 6 + × 6 = 2. The central atoms of upper and lower planes are 6 6 shared by two unit cells which means that upper and lower planes contribute 1/2 atom each. Hence the

with reference to corner atoms is

1 × 2 = 1 and there are three interstitial 2 atoms. Thus the total number of atoms in hcp unit cell = 2 + 1 + 3 = 6. The volume of the atoms in the 4 unit cell, v = 6 × πr 3 . To calculate the volume of the unit cell V one requires the relation between the 3 height c and the edge a assuming that the atoms in the top and the bottom layers touch each other along the edges of the unit cell, with the relation a = 2r.

contribution of central atoms in both upper and lower planes is

C

a Fig. 1.16 Ideal hexagonal close packed structure

Computation of

c ratio: a Z 0

c/2 a

A

A

o a B

30º

Y

X

a a B

Fig. 1.17 Sketch of the bottom layer

From D AYB,

cos 30° =

AY a 3 ; AY = AB cos 30° = AB 2

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57

Also in the triangle AOB, 2 2  a 3  a = ( AY ) =  3 3  2  3 From DAZX, (AZ)2 = (AX)2 + (XZ)2

AX =

a2  c  +  = 3 2

i.e.,

2

a2 c2 + 3 4 c2 2a 2 = 4 3 a2 =

c = a

or

8 3

...(1.56)

Volume of the unit cell is obtained as follows: a 3 1 1 Area of the triangle AOB = ( BO) ( AY ) = a × 2 2 2

Hence area of the base = 6 × Thus volume of the unit cell

a2 3 4

= 6×

3 a2 × c 4

or V=

3 3 a2c 2

Now APF = APF = Substituting a = 2r and

v V 6 × (4 / 3) π r 3 (3 / 2) 3 a 2 c

a 3 = and simplifying, we get c 8 APF =

π 3 2

APF = 74%

= 0.74 ...(1.57)

Based on the value of packing factor it may be seen that sc is loosely packed structure, bcc is known as closely packed structure and fcc and hcp are closest packed structures. The details of coordination number, atomic packing factor etc. for cubic and hexagonal close packed structures are given in Table 1.3. Structures of some important solids and lattice constants are given in Table 1.4.

58


Table 1.3 Comparison of cell properties of some crystal lattices Sl. No

Simple cube (sc)

Properties

Body centred cube (bcc)

Face centred cube (fcc)

Hexagonal close packed structure (hcp)

1.

Volume of unit cell

a3

a3

a3

3 3 (a 2 c) 2

2.

Number of atoms per unit cell

1

2

4

6

1

2

4

4

3.

Number of atoms per unit volume

3

3

3

a

a

a

3 (a 2 c)

4. Number of nearest neighbours

6

8

12

5.

Nearest neighbour distance (2r)

a

a 3 2

a 2 2

6.

Atomic radius

a 2

a 3 4

a 2 4

a 2

π 3 = 0.52 6

π 3 = 0.68 8

π 2 = 0.74 6

π 2 = 0.74 6

Sodium, Iron Aluminium, and Chromium Copper and Lead

Magnesium, Zinc and Cadmium

7. Atomic packing factor 8.

Examples

Polonium

12

a

Table 1.4 Structures of some important solids Solids

Crystal structure

Lattice constants (nm)

Aluminium Chromium Cobalt

fcc bcc hcp

0.405 0.288 a = 0.251 c = 0.407

Copper Iron Magnesium

fcc bcc hcp

0.361 0.287 a = 0.321 c = 0.521

Nickel Zinc

fcc hcp

a = 0.352 a = 0.266 c = 0.495

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59

1.15 CRYSTAL DIRECTIONS

Crystal Directions It is always useful to have a convention or standardized procedure to specify the directions in a crystal. The procedure of finding the directions inside the crystal is explained below. 1. Consider any lattice point that lies on the line as origin. 2. Write down the position vector of the next nearest point on the line in terms of the fundamental → →

→

translation vector a , b and c of the unit cell of the crystal, say, →

→

→

→

r = r1 × a + r2 × b + r3 × c →

3. Now the components of position vector r along the three directions of a, b, c are r1, r2, r3 respectively. Then the crystal direction is denoted by [r1r2r3]. Let us apply this procedure to find the directions of OP, OQ and OR of an orthorhombic unit cell (a ¹ b ¹ c; a = b = g = 90°) in Figure 1.18 taking ‘O’ as origin. a. Direction of OP →

→

→

Position vector of OP = 1× a+1×b +1× c OP = \ r1 = 1; r2 = 1; r3 = 1 \ Direction of OP is represented by [111] b. Direction of OQ →

→

→

Position vector of OQ = 1× a+0×b +0× c OQ = \ r1 = 1; r2 = 0; r3 = 0 \ Direction of OP is represented by [100] c. Direction of OR →

→

→

1× a+0×b +1× c Position vector of OR = OR = \ r1 = 1; r2 = 0; r3 = 1 \ Direction of OP is represented by [101] It should be understood that the directions [222],[333],[444], ……..[nr1 nr2 nr3] will all coincide with [111]. In such cases the lowest combination of integers i.e., [111] is used to specify the direction. If any of the integers is negative, for example –3, it should be written as 3 which is read 3 bar. Given three integers of a direction, a family of directions with different possible combinations of them, both positive and negative, is represented with brackets of the type < >. For example, a family of directions such as [132], [312], [123], [1 3 2], [ 1 32], [ 1 2 3 ], [ 1 3 2], [ 1 3 2 ] etc., is represented by <132>.

60


z

R

P [ 1 1 1]

[1 0 1]

c

y 0 Q

[1 0 0]

a

b x

Fig. 1.18 Crystal directions

1.16 CRYSTALLOGRAPHIC PLANES AND MILLER INDICES As atoms are regularly grouped in crystal, it is necessary to name the crystallographic planes or atomic planes in which the atoms are arranged in a periodic manner. These planes and crystal directions play an important role in hardening reaction, plastic deformation and other properties of materials. X-rays are easily diffracted by planes containing large concentration of atoms and hence a knowledge of such planes is essential in structure analysis. Miller evolved a method to designate a set of equidistant parallel planes in a crystal by three numbers (h k l ) known as Miller indices. Miller indices are a set of integers which represent a set of equidistant parallel planes.

1.16.1 Method of Finding Miller Indices The steps involved in the determination of Miller indices of a set of equidistant parallel planes are illustrated with the aid of Fig. 1.19. Consider any one plane from the set of equidistant parallel planes. Then the following procedure is adopted to find the Miller indices. (i) Determine the co-ordinates of the intercepts made by the plane along the three crystallographic axes (x, y, z). 2a, 3b, c (ii) Express the intercepts as fractions of the unit cell dimensions along the axes, i.e., c 2a 3b a b c 2 3 1 (iii) Get the reciprocals of these numbers. 1 1 1 2 3 1 (iv) Reduce these reciprocals to the smallest set of integers and enclose them in brackets without any punctuation between them. They are the Miller indices of the given set of equidistant parallel planes. 1 1 1 6× 6× 6× 2 3 1 3 2 6 (3 2 6) ...(1.58)

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61

Z

3c 2c c

0

b 2 b 3b 4b

Y

a 2a 3a X Fig.1.19 Miller indices of an important plane

CONCLUSIONS (i) All the parallel equidistant planes have the same Miller indices. Thus Miller indices define a set of equidistant parallel planes. (ii) A plane parallel to one of the coordinate axes has an intercept of infinity on that axis. (iii) If the Miller indices of two planes have the same ratio, i.e., (8 4 4) and (4 2 2) or (844) and (2 1 1), then the planes are parallel to each other. In other words, planes with Miller indices (nh nk nl) and (h k l) will be parallel to each other. (iv) If (h k l) are the Miller indices of a plane, then the plane cuts the a, b, c – axes into h, k and l equal parts respectively.

1.16.2 Miller Indices of Some Selected Planes Let us now obtain the Miller indices of the planes (a) BDK, (b) BDHF, (c) CDHG, (d) ABFE and (e) ACH with the help of the Figs. 1.20 (a) to (e). (a) Plane BDK Axes x y z Coordinates of the intercepts

a

b

Fraction

1

1

1 (112)

1

c 2 1 2 2

x a 1 1 (110)

y b 1 1

z ¥ ¥ 0

Reciprocal Miller indices of plane BDK are (b) Plane BDHF Axes Coordinates of the intercepts Fraction Reciprocal Miller indices of plane BDHF are

62


Z

Z

E

H

F

E

G

H

F

G

K

D

A B

Y B

C

X

Z

F

D

A

H

E

G

F

(b )

Z

X H

G

D

A

Y

B

B

C

(d ) X

Z

Z

H

E

B

C

G

F

D

A

H

E

G

F

D

A

Y B

(e ) X

Y

C

(c )

X

Y

C

(a )

E

D

A

C (f)

X Fig.1.20 Determination of Miller indices

Y

I


(c) Plane CDHG Axes Coordinates of the intercepts Fraction Reciprocal Miller indices of plane CDHG are

63

x ¥ ¥ 0 (010)

y b 1 1

z ¥ ¥ 0

(d) Plane ABFE The plane ABFE passes through the origin. Hence the Miller indices of the plane ABFE cannot be determined without changing/shifting its origin. Now any point of the unit cell not lying on the plane may be selected as origin. Let the point D be chosen as the origin. It should be noted that the plane ABFE is parallel to X and Z axes and it intersects the Y axis on negative side. We shall now follow the regular procedure to find the Miller indices of the plane. Axes x y z Coordinates of the intercepts ¥ –b ¥ Fraction ¥ –1 ¥ Reciprocal 0 –1 0 Miller indices of plane ABFE are (0 1 0) (Note : 1 is read one bar) (e) Plane ACH The plane ACH passes through the origin. Let us choose D as origin and follow the regular procedure to find the Miller indices of the plane ACH. Axes x y z Coordinates of the intercepts a –b c Fraction 1 –1 1 Reciprocal 1 –1 1 Miller indices of plane ACH are (1 1 1) (f) Plane BDG The Miller indices of plane BDG can be found by taking the point C as origin. Axes x y z Coordinates of the intercepts –a –b c Fraction –1 –1 1 Reciprocal –1 –1 1 Miller indices of plane BDG are ( 1 1 1)

1.17 DETERMINATION OF INTERPLANAR DISTANCE Let us assume that we are interested in estimating the interplanar distance of a set of equidistant parallel planes of Miller indices (hkl) in a cubic unit cell. Let ABC be the plane (Fig. 1.21) with Miller indices (hkl) nearest to the origin ‘O’. Let PQR be the next nearest plane to the origin with Miller indices (hkl). Let ON be the perpendicular drawn to the plane ABC from the origin. When extended, it intercepts PQR at N'. Let a, b, g be the angles made by ON with x, y and z axes respectively. (Please note that in this discussion a, b, g are not unit all parameters).

64


We know that the intercepts of the plane ABC on the three axes x, y and z are OA, OB and OC respectively. Now a a a OA = , OB = and OC = where ‘a’ is the side of the cubic unit cell. l k h dl dk d dh ON and cos γ = 1 where d1 is the length of ON or the = 1 = 1 , cos β = 1 cos α = a a OA a / h a perpendicular distance of ABC from the origin. Since cos2 α + cos2 β + cos2 γ = 1 , we have, 2

2

2

 d1 h  d k d l   + 1  + 1  = 1  a   a   a 

or

∴ d1 =

d12 (h 2 + k 2 + l 2 ) = a 2

a

h + k2 + l2 In a similar way we can show that the distance of the next parallel plane having the same Miller indices from the origin as 2

2a

d2 =

where ON' = d2.

h2 + k 2 + l2

\ The interplanar distance of planes with Miller indices (hkl) is

a

d = d2 – d1 =

h + k 2 + l2 2

a Thus

d=

...(1.59)

h2 + k 2 + l2 z

z R

M

M C

C B

B

Y

0

N'

N

Q

0

A

A x

z

P

x

N d1 γ O

α

β y

x Fig. 1.21 Interplanar spacing in cubic crystal

y

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65

PROBLEMS AND SOLUTIONS IN STRUCTURE OF SOLIDS 1.3.1 Calculate the number of atoms per unit cell of a metal having lattice parameter 0.29 nm and density 7870 kg/m 3. Atomic weight of the metal is 55.85 and Avogadro’s number is 6.02× 1026/kmol. (Anna University, B.E., December 1998, JNTU, B.Tech., April 2001) Solution The general expression for the lattice constant is a3 =

n=

=

n MA (Refer solution of SCP3) NA ρ

ρ N A a3 MA 7870 × 6.02 × 10 26 × a 3 55.85

7870 × 6.02 × 10 26 × 0.293 × (10 −9 ) 3 55.85 = 2.07 n=2 Answer. 1.3.2 Since the number of atoms per unit cell which is equal to the number of lattice points per unit cell is 2, the given metal should have bcc structure. Compute d200/d111 in nickel in fcc structure. The radius of the Ni atom is 0.1245 nm. (Anna University, B.E., April 1995, Madras University, B.E., April 1992). Solution n=

a= d 200 =

d111 =

4r 2

=

4 × 0.1245 = 0.352 nm 1.414

a 0.352 = = 0.176 nm 2 2 a 3

=

0.352 3

= 0.203nm

 a a  d200 = =    22 + 02 + 0 2 2    a a   d111 = =  3  12 + 12 + 12 

d 200 0.176 = d111 0.203 d 200 = 0.867 d111

Answer.

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1.3.3 The lattice constant of aluminium (f c c) is 1.4 times the lattice constant of chromium (b c c) structure. Compare the radii of the atoms of these metals. (REC Calicut, B.E., December 1997). Solution 4rA aA = 2 4rc ac = 3

aA rA ac = rc

3 2

1.4 ac rA = ac rc

3 2

1.4 rc × 1.414 1.732 rA = 1.14 rC Answer. rA =

OBJECTIVE QUESTIONS IN STRUCTURE OF SOLIDS 1. In a unit cell of Fe and Cu the ratio of percentage of volume of unit cube occupied by the atoms is (a) 0.92 (b) 0.74 (c) 0.68 (d) 1 2. The lattice constant of a bcc unit cell is 0.2864 nm. The nearest neighbour distance is (a) 0.54 nm (b) 1.2 nm (c) 0.496 nm (d) 0.248 nm 3. A plane in a unit cell is described by its Miller indices (6 3 2). The coordinates of intercepts of the plane (632) nearest to the origin are (a) 6, 3 and 2 units (b)

a a a , and 6 3 2

(c)

a a , and a 3 3

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1 2 , 1 and units 2 3 4. A fixed mass of pure iron cools from 1000°C to room temperature. Its volume will increase because (a) the radius of Fe atom increases at lower temperature (b) the iron is in fcc at 1000°C and changes to bcc at room temperature (c) the distance between iron atoms increases at lower temperature (d) the iron in bcc structure at 1000°C changes to fcc at room temperature 5. In a bcc unit cell the central atom is in contact with (a) four identical atoms (b) six-identical atoms (c) twelve identical atoms (d) eight identical atoms

(d)

SHORT QUESTIONS IN STRUCTURE OF SOLIDS 1. Explain the terms: space lattice and translation vector. 2. Distinguish between space lattice and crystal structure. 3*. Show that for a cubic lattice, the lattice constant a is given by

n MA  a=    NA ρ 

4. 5. 6*. 7. 8.

1/ 3

where n is the number of atoms in the unit cell, MA is the atomic weight, NA is Avogadro's number and r is the density of the metal. Distinguish between primitive cell and unit cell. Explain co-ordination number, nearest neighbour distance and packing factor with the help of a crystalline solid. Give the general explanation of closest packing with suitable figures. What are Miller indices? How are they obtained? Write down the nearest neighbour distance, co-ordination number and packing factor of hcp structure.

ANSWERS TO THE STARRED QUESTIONS

3. Let Now

MA ρ represent the molar volume of a crystal, say, any metal . MA ρ will contain NA number of atoms.

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 N A ρ a3   number of atoms. Therefore the volume a3 will contain  M A   This must be equal to the number of atoms in the unit cell, say, n. Thus

N A ρ a3 =n MA or

n MA  a=   ρ NA 

1/ 3

Answer.

6. Closest packing is a way of arranging equidimensional objects such that the available space is more efficiently filled. Under this condition each object will be in contact with the maximum number of like objects. A R

B C (a)

(b )

The co-ordination number in the two dimensional case is 6 and that in the three-dimensional case is 12.

REVIEW QUESTIONS IN STRUCTURE OF SOLIDS 1. Explain the terms: Lattice, Basis, Structure and Unit cell. Estimate the fraction of total volume occupied by atoms in body centred and face centred cubic structures. Mention two elements each crystallizing in these structures (Madras University, B.E., November 1998, Anna University, April 1990) c 2. Show that in ideal hexagonal close packed structure ratio is 1.663 and the density of a atomic packing factor equals to that of the face-centred cubic structure. (Anna University, B.E., November 2000) 3. Explain how Miller indices of a set of parallel planes in a crystal are determined. Show that for a cubic crystal the distance dhkl between adjacent (h k l) planes is given in terms of the unit cell edge a by a d hkl = (JNTU, B.E., April 2000) 2 h + k 2 + l2

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EXERCISE IN STRUCTURE OF SOLIDS 1 Aluminium has a density of 2698 kg/m3. Its structure is fcc and atomic weight is 26.98. Calculate (a) How many atoms are contained in 1m3 of this metal? (b) Get the size of the unit cube for aluminium. (c) Calculate the atomic radius of aluminium. (JNTU, B.E., December 2001) (Ans: 6 × 1028, 6.7 × 10–29m3, 0.176 nm) 2. The toy-car of Mr. Ashok was heavily damaged; he threw all the parts except the chassis. It is of a-iron with bcc structure at 300K and it is a rectangular plate with dimensions 10 cm, 8 cm and 1 cm. Compute the number of unit cells in this specimen. The nearest neighbour distance of this structure is 0.244 nm. (Ans: 3.57 × 1024) 3. In a crystal whose primitives are 0.12 nm, 0.18 nm and 0.2 nm. A plane (2 3 1) has an intercept of 0.12 nm on the X-axis. Find the intercepts on the Y and Z axes. (Bangalore U, B.E., May 1982) (Ans: 0.12 nm, 0.4 nm)

Unit 01A - New Age International

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