Worked solutions Chapter 15 Sound - PEGSnet

Worked solutions 

Chapter 15 Sound 

15.1 The nature of sound 

1 C. Sound is a mechanical wave form, energy is transferred by the molecules within a medium. 

2 C and E both apply. 

3 A. Sound is a longitudinal wave form. 

4 B. Energy is transferred away from the tuning fork. 

5 In a transverse wave, particles vibrate at right angles to the direction of travel. The vibration is 

in the same direction as the direction of propagation. 

6 From scale, distance from compression at P to compression at R is 0.60 m. 

7 a Microphones convert sound energy to electrical energy. 

b From graph supplied maximum pressure variation is at both peaks of the graph. 

so t = 0.5, 1.5, 2.5, 3.5, 4.5 and 5.5 ms. Maximum pressure variation is an absolute value, 

so sign isn’t considered. 

8 a Maximum positive pressure variations are experienced at compressions. 

b Maximum negative pressure variations are experienced at rarefactions. 

c At points midway between compressions and variations, pressure variation is zero. 

d At points midway between compressions and variations, pressure variation is zero. 

9 a Maximum positive pressure variations are experienced at compressions; 

hence, t = 0.5, 2.5, 4.5 ms. 

b Maximum negative pressure variations are experienced at rarefactions; 

hence, t = 1.5, 3.5, 5.5 ms. 

Heinemann Physics 12(3e) 

Copyright © Pearson Education Australia 2008 

Teacher’s Resource and Assessment Disk 

(a division of Pearson Australia Group Pty Ltd) ISBN 9781442501263 

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15.2 The wave equation 

1 C. The speed of sound is the speed at which energy is being transferred by the sound wave. 

2 There is a greater force of attraction between the molecules in water than those in air. When a 

water molecule is displaced, it thus returns to the mean position faster, transferring energy 

faster. 

3 C. A greater air temperature means a greater energy level overall. 

4 D is the best answer since amplitude is a measure of the energy. However, it could also be 

argued that as A, B and C are to a degree interdependent, the other choices are also correct. 

5 C. Amplitude is independent of the other quoted quantities. 

0.30 

6 See page 556 for relevant figures. v = d/t = 

3 

1" 

10 ! 

= 300 m s –1 

7 t for sound to travel 20 m = 

Hence advantage = 0.059 s 

Moral—watch the gun! 

20 

340 

= 0.059 s, time for light to travel 10 m is negligible. 

8 Total time = 20 s = time for shell to travel to target plus time for sound to travel back. 

Time for sound to travel back = 3400 = 10 s; hence, time for shell to reach target = 10 s and 

340 

speed of the shell = 340 m s –1 . 

9 a v = f , hence = v/f = 1500 = 1.5 m. 

1000 

b Total time = time to reach floor plus time to come back. 

Hence, time to floor = 1.0 s and distance = 1.0 × 1500 = 1500 m or 1.5 km. 

c The medium does not affect the frequency, hence ratio = 1:1. 

10 a = v/f = 1500 

2000 = 0.75 m 

(Doubling frequency halves wavelength since f is inversely proportional to .) 

b The medium is unchanged. Hence, time remains the same. 





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15.3 Diffraction of sound 

1 = v/f = 340 

1000 = 0.34 m 

2 Since is approximately equal to w, appreciable diffraction occurs and the sound will spread to 

points P and R and also be heard at Q. 

3 = v/f = 

340 

4000 = 0.085 m 

4 is significantly smaller than w. Hence, less diffraction will occur. 

5 More of the sound energy travels directly to Q. 

6 B. An increase in amplitude means more energy overall, so an increase in intensity will be 

detected. 

7 D. With an increase in W, diffraction will decrease. 

8 High frequencies are diffracted less than low frequencies and, hence, are more directional. 

9 A. The wavelength of the sound from the double bass is longer; hence, will be diffracted more. 

10 Since the frequency of the violin is higher, the wavelength will be shorter. Less diffraction will 

occur, meaning that more of the available sound energy will reach the teacher at point Y. 





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15.4 Amplitude, intensity and the decibel scale 

1 D. Turning up the volume increases energy and therefore increases intensity. It has no effect on 

the other quantities. 

2 The frequency response of any two individual’s ears will be different and hence the individuals 

may differ on the subjective quantity of a sound’s loudness. This doesn’t mean the quantifiable 

energy level is different. 

3 E. Power output, distance, medium and prevailing conditions will affect the intensity of the 

sound received. 

4 a = 340 = 0.23 m or 23 cm 

1500 

b = 23 cm since the wavelength of the sound source won’t change. It is not dependent on 

the distance from the source. 

5 a Distance source to P = 1 m, to Q = 3 m. 

1 

Since I ∝ 

2 

r , I = 1 

= 4.4 × 10 –3 W m –2 

2 

32" 4.0" 

10 ! 

1 

b Distance is now 5 times original, so I = 

2 

52" 4.0" 

10 ! 

= 1.6 × 10 –3 W m –2 

6 a 

1 

I = P/A = 

2 

4 ! ! ! r 

b 

1 

I = P/A = 

2 

4 ! ! ! r 

1 

= 

113 = 8.8 × 10–3 W m –2 

1 

= 

314.2 = 3.2 × 10–3 W m –2 

7 E = Pt = 1 × 10 = 10 J 

8 a Using L = 10 log I/I o where I o = 10 –12 W m –2 , and I = P/A = 7.9 × 10 –2 W m –2 , L = 109 dB 

b Using L = 10 log I/I o where I o = 10 –12 W m –2 , and I = 8.8 × 10 –3 W m –2 , L = 99 dB 

c Using L = 10 log I/I o where I o = 10 –12 W m –2 , and I = 1.6 × 10 –3 W m –2 , L = 95 dB 

9 a Using L = 10 log I/I o where I o = 10 –12 W m –2 , and L = 100 dB, I = 1.0 × 10 –2 W m –2 

1 

b At Q, I = = 1.1 × 10 –3 W m –2 

2 

32" 1.0" 

10 ! 

c Using L = 10 log I/I o where I o = 10 –12 W m –2 , and I = 1.1 × 10 –3 W m –2 , L = 90 dB 

10 An decrease in sound level of 3 dB corresponds approximately to a halving of the sound 

intensity. 12 dB would therefore mean the sound intensity was halved 4 times, so 

I = 5.0 × 10 –10 × 24 = 8.0 × 10 –9 W m –2 

11 a 

1 

I = = 1.0 × 10 –5 W m –2 

2 3 

10 " 1.0" 

10 ! 

b Using L = 10 log I/I o where I o = 10 –12 W m –2 , and I = 1.1 × 10 –5 W m –2 , L = 70 dB 





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15.5 Frequency, perceived loudness and the phon 

1 Frequency is determined by the source of the sound and is independent of the listener. Pitch is 

how the frequency of the sound is perceived by a particular listener. 

2 D. Average hearing for a young person is in the range of 20 Hz to 20 kHz. 

3 C. Hearing loss due to age is usually confined to higher frequencies. 

4 A is the better answer within the context of this question. Higher frequencies do carry more 

energy; hence, B is correct but not within the question’s context. 

5 Referring to the phon curves on page 574: 

a 0.2 kHz or 200 Hz 

b 0.1 kHz or 100 Hz 

c 0.05 kHz or 50 Hz 

6 Referring to the phon curves on page 574, i.e. 90 dB. 

7 Referring to the phon curves on page 574, i.e. 400–2000 Hz. 

8 C. Following the 60dB phon curve to where it intersects with a sound level of 80dB, i.e. 50Hz. 

9 a P = IA = 1.0 × 10 –8 W m –2 × (4 × π × 12) = 1.26 × 10 –7 W 

b From the phon curve I = 1000 × 10 –12 W m –2 or 1.0 × 10 –9 W m –2 . 

Using L = 10 log I/I o where I o = 10 –12 W m –2 , and I = 1.0 × 10 –5 W m –2 . L = 30 dB (based 

on the graph as printed. The published answer requires more careful scaling to achieve an 

answer of 25 dB.) 

10 At 10 kHz I = 10 –8 W m –2 , and at 2.0 kHz I = 10 –12 W m –2 . 

Using L = 10 log I/I o where I o = 10 –12 W m –2 , and I = 1.0 × 10 –8 W m –2 . L = 40 dB 





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15.6 Making sound: strings and air columns 

1 Refer to the theory of superposition of waves to confirm these answers. 

a true 

b false 

c true 

d false 

2 When the glass is exposed to a sound of the same resonant frequency as its natural frequency of 

vibration, resonance will occur. The amplitude of the vibrations will increase. If sufficient 

energy is available, the glass may shatter. 

3 The sound box of a guitar is tuned to resonate in the range of frequencies being produced by the 

guitar strings. Resonance within the sounding box amplifies the sound. 

4 A standing wave is produced as a result of the superposition of two waves of equal amplitude 

and frequency travelling in opposite directions in the same medium. 

5 a C. Refer to graphs page 586. 

b B. Refer to graphs page 586. 

c C. Refer to graphs page 586. 

6 a At the centre, i.e. L = / 2 

b At a point one-quarter of its length, i.e. L = / 4. 

c At a point one-sixth of its length, i.e. 2L = / 3 or / 6. 

7 a = 2L = 1.0 m, f = v/ = 300 Hz 

b f 2 = 2 × f 1 = 600 Hz 

c f 3 = 3 × f 1 = 900 Hz 

8 Resonance in air columns of a particular length is due to reflection of waves arriving at the ends 

of the column. The reflected waves are superimposed on the existing waves to produce a 

standing wave pattern. This results in resonance. 

9 a 1= 2L = 90 cm or 0.90 m 

b 2= 2L/2 = 45 cm or 0.45 m 

c 3= 2L/3 = 30 cm or 0.30 m, f = v/ = 330/0.30 = 1100 Hz or 1.1 kHz 

10 a f 1 = v/ = 330/2L = 110 Hz 

b f 3 = 3 × f 1 = 330 Hz 

c Only odd numbered harmonics are possible; 

so f 5 = 3 × f 1 = 550 Hz and f 7 = 3 × f 1 = 770 Hz 





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15.7 Recording and reproducing sound: the first and last links 

1 A moving coil loudspeaker typically uses a flexible cone to produce sounds, while a moving 

coil microphone uses a shielded diaphragm. The flat surface of the diaphragm provides a more 

reliable surface for converting sound waves to an electrical current. 

2 The larger the loudspeaker, the larger the cone and the more difficult it would be to make it 

vibrate with incident sound waves. Also, the 3D nature of the cone would make accurate 

reproduction of incident sound waves difficult. Thus a small, relatively flat cone—typical of 

small mid-range or tweeter speakers—would be the most suitable. 

3 The small, flat cone of a tweeter could not vibrate at the low speeds necessary to reproduce low 

frequencies reliably. 

4 Dropping the cone onto a bench would cause the coil of wire glued to the cone to move within 

the magnetic field of the magnet. An induced voltage would result and would be seen as a jump 

in the trace of an oscilloscope. 

5 A cylindrical magnet produces a strong radial field in the gap between north and south poles. 

The coil of the speaker vibrates within this field. The stronger the magnetic field, the higher— 

and therefore more distinct—the induced current/voltage. 

6 Their frequency range is too limited. (Carbon microphones cannot respond to frequencies above 

4 kHz.) 

7 The combined frequency responses of the different speakers need to cover the full range of 

frequencies audible to humans, while accurately producing particular small groups of 

frequencies. 

8 The electrodynamic, or moving coil, microphone provides the broad frequency response 

required for recording music. 

9 A small diaphragm is attached to a coil of wire. When sound energy is incident on the 

diaphragm, it moves up and down, moving the coil between the poles of a permanent magnet. 

This generates an induced voltage of varying frequency and amplitude in the coil, directly 

related to the variation in sound energy incident on the microphone’s diaphragm. 

10 As both microphones and loudspeakers convert one form of energy to another, both could be 

accurately described as transducers. 





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Chapter review 

1 C. Based on the graphs supplied pressure, variation is the best answer. 

2 B. From amplitudes of graphs supplied. 

3 C. From horizontal scale of graphs supplied. 

4 A. There is only one complete wave hence only one rarefaction. 

5 C. The graph of pressure variation has three positive peaks corresponding to three compressions 

(rarefactions are represented by negative peaks). 

6 A. Pressure variation is a maximum midway between a compression and rarefaction. 

7 C. One complete wave along the horizontal scale. 

. 

8 A. The wave is moving away from the source from left to right. 

9 D. From graph supplied. 

10 C. Speed is determined by the medium. 

11 Both C and D are correct. 

12 Using v = 340 m s –2 , f 1 = v/4L = 

13 f 3 = 3 × f 1 = 300 Hz 

340 

4! 

0.85 

= 100 Hz 

14 P = IA = 1.6 × 10 –6 W m –2 × (4 × π × 10 2 ) = 2.0 × 10 –3 W 

1 

15 At 20 m the distance is doubled so I = 

6 

22" 1.6" 

10 ! 

W m –2 = 4.0 × 10 –7 W m –2 

16 E = Pt = 2.0 × 10 –3 W × 60 × 10 = 1.2 J 

17 Intensity of sound is a measurable quantity (units W m –2 ), while the loudness of a sound 

depends on how a particular individual’s ears respond to that sound and can vary with frequency 

for each individual. 

18 An increase of 3.0 dB in sound level represents a doubling in sound intensity. Prolonged 

exposure to this increased level could cause problems. 

19 Most participants will report the greatest differences in perceived loudness at frequencies below 

500 Hz and above 5.0 kHz, i.e. at the lower and upper limits of average human hearing range. 

20 A tweeter best produces frequencies between 4 and 20 kHz. It would distort sounds, and test 

results, for frequencies significantly below 4 kHz. For best results, a broad range speaker with a 

flat frequency response curve for frequencies between 20 Hz and 20 kHz (the range of normal 

human hearing) would be required. 





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21 Baffling reduces, but rarely eliminates, reflections of sound waves that are created by the 

movement of the back surface of a loudspeaker’s cone. If not reduced, these sound waves would 

be superimposed on the sound waves from the front of the speaker cone to cancel out, or reduce, 

some frequencies. 

22 Electret-condensor. Mobile phones need to reliably reproduce the frequency range of the human 

voice—generally below 1 kHz. The microphone also needs to be compact and durable. The 

small capsule, low cost and durability of an electret-condensor microphone would be a good 

option. Its poor response to high frequencies would not be a drawback for this application. 





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Worked solutions Chapter 15 Sound - PEGSnet

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