Chapter 13

C Waves Chapter 13 Sound

13 Sound

Practice 13.1 (p. 193)1 B

By v = f,

= = 0.113 m

Since the separation of the loudspeakers

should be of a few wavelengths, 0.3 m is the

most appropriate separation.

2 A

3 C

4 (a)

(b) Since low frequency noise has longer

wavelength, it bends more when passing

the barrier. Hence, flats in the shadow

of the barrier are still affected by the

low frequency noise and barrier is not

effective to shield the building from

such noise.

5 (a) By v = f,

= = = 0.68 m

= 2.21 wavelengths

The doorway of 1.5 m is 2.21

wavelengths wide.

(b) Diffraction is a property of sound

waves. Since the width of the doorway

is comparable to the wavelength of the

sound waves, waves bend around the

doorway when they pass through it.

6 (a) By v = f,

= = = 0.34 m

The two loudspeakers are 1 m apart.

= 2.94 wavelengths

Therefore, they are 2.94 wavelengths

apart.

(b) If the two loudspeakers are 0.5 m apart,

the loud sounds in the interference

pattern become more widely spaced.

7 If the frequency of the speakers were

lowered, the wavelength of the waves

increases. The nodal and antinodal lines of

the interference pattern decrease in number

and are more widely spaced. Therefore, the

points C and D move farther apart.


2 D

3 B

4 C

Let a be the distance between the boy and

cliff A and b be the distance between the boy

and cliff B.

The first echo is heard at t = 0.8 s.

s = vt

New Physics at Work (Second Edition) Ó Oxford University Press 2006123


2a = 340 0.8

a = 136 m

The second echo is heard at 0.88 s later.

s = vt

2b = 340 0.88

b = 149.6 m

Hence, distance between the cliffs

= a + b = 286 m

5 By v = f,

= = = 1.3 m

Therefore, the wave compressions are 1.3 m

apart. The wave rarefactions are also 1.3 m

apart.

6 By v = f,

= = = 0.833 m

The wavelength of the sound wave travelling

along the iron rod is 0.833 m.

7 s = v t = 333 = 249.75 m

The length of the lake is 249.75 m.

8 (a) (i) By s = vt,

t =

= = 5.88 s

The time taken to hear the bang is

5.88 s.

(ii) By s = vt,

t =

= = 6.67 10–6 s

The time taken to see the flash

from the cannon is 6.67 10–6 s.

(b) From (a)(i) and (a)(ii), the ratio of the

times

=

= 882 000 : 1

9 – = 1.4

s = 511 m

The impact occurred at 511 m away.

10 v = st = 340 2.5 = 850 m

The firework is 850 m from the audience

when it explodes.

11 (a) By v = f,

= = = 0.01 m

The wavelength of the ultrasonic waves

emitted is 0.01 m.

(b) A small wavelength results in a less

degree of diffraction when the

ultrasonic waves encounter any

obstacles. Therefore, ultrasound can be

used to detect the specific targets more

accurately.

(c) s = vt = 1500 = 75 m

The shoal of fish is 75 m below the ship.

12 s = vt = 340 = 34 m

The distance between the bat and the bird is

34 m.

13 The time period between wing beats

=

=1.667 10–3 s

s = vt

= 340 1.667 10–3

= 0.567 m



The sound waves travel 0.567 m between

wing beats.

14 v =

= = 333 m s–1

The speed of sound is 333 m s–1.

15 (a) By v = f,

f =

=

= 255.6 Hz

v = f

= 255.6 5.47

= 1400 m s–1

The speed of sound in the medium is

1400 m s–1.

(b) This phenomenon is called refraction.


2 D

3 D

4 (a)

(b)

(c)

(d)

5 (a) T =

= = 3.83 10–3 s

The period of one vibration of this tone

is 3.83 10–3 s.

(b) By v = f,



=

=

= 1.3 m

The wavelength of the note is 1.3 m.

6 (a) Noises at 130 dB produce ear pain.

(b) Students’ conversation and

loudspeakers are two sources of noise in

school.

(c) Students talk gently and lower the

volume of loudspeakers.

Revision exercise 13Multiple-choice (p. 218)Section A

1 B

2 B

3 C

4 C

5 D

6 (HKCEE 2003 Paper II Q29)


Section B

8 C

The path difference

= S2P – S1P = 4.2 m – 3.4 m = 0.8 m

Constructive interference takes place at a

position where the path difference is equal to

whole number of wavelengths. Therefore, it

is impossible that the wavelength of the

sound equals to 0.6 m.

9 D

10 D




Conventional (p. 220)Section A

1 (a) Assume the time of travel by light is

negligible.

v = (1M)

= = 345 m s–1 (1A)

The speed of sound is 345 m s–1.

(b) s = vt (1M)

= 345 0.2 = 69 m (1A)

The height of the cliff is 69 m.

2 (a) The elastic string sets the air particles

around it to vibrate. (1A)

The vibrations of air particles are

transmitted to Joe’s ear and hence he

can hear the sound. (1A)

(b) (i) f = (1M)

=

= 500 Hz (1A)

The frequency of the sound

produced is 500 Hz.

(ii) By v = f, (1M)

=

=

= 0.68 m (1A)

The wavelength of the sound is

0.68 m.

3 (a) (i) When the loudspeaker produces a

sound, the loudspeaker cone

moves in and out rapidly. (1A)



This stretches and compresses the

air in front. (1A)

As a result, a series of rarefaction

and compression travels through

the air and the flame follows the

motion of air. (1A)

(For effective communication.)

(1C)

(ii) By v = f, (1M)

=

=

= 0.85 m (1A)


0.85 m.

(b) Any two of the following reasons:

(2 1A)

The frequency of the signal generator is

out of the audible range.

The amplitude of sound is too small.

There is a vacuum in the box.

4 (a) The frequency range of sound given by

hummingbirds is from 15 Hz to 80 Hz.

(1A)

(b) By v = f, (1M)

= = = 22.7 m (1A)

The longest wavelength of the humming

sound is 22.7 m.

By v = f,

= = = 4.25 m (1A)

The shortest wavelength of the

humming sound is 4.25 m.

(c) No, human cannot hear the whole range

of sound made by hummingbirds. (1A)

This is because audible frequency

ranges from 20 Hz to 20 kHz only. (1A)

5 (a) Get a partner to hit a long iron rail from

a distance. (1A)

Then hear the sounds travelling through

first the rail and then the air. (1A)

(b) t =

=

= 29.4 s (1A)

t =

=

= 2.5 s (1A)

The time difference

= 29.4 s – 2.5 s = 26.9 s (1A)

Therefore, when compared with those

soldiers standing up and listening, the

soldier can detect the enemy earlier by

26.9 s.

6 (HKCEE 2003 Paper I Q5)


Section B

8 (a) (i) S1Q = (1M)

= 8.14 m (1A)

The distance S1Q is 8.14 m.

(ii) S2Q = (1M)

= 8.38 m (1A)

The distance S2Q is 8.38 m.

(b) The path difference of Q

= S2Q – S1Q (1M)



= 8.38 m – 8.14 m

= 0.24 m (1A)

(c) The path difference of Q =

The wavelength of the sound emitted is

0.24 m. (1A)

(d) v = f (1M)

= 1.4 1000 0.24

= 336 m s–1 (1A)

The speed of sound in air is 336 m s–1.

9 (a) Yes, notes X, Y and Z have the same

pitch. (1A)

It is because the CRO settings for the

displays are the same and all displays

have the same number of waveforms,

i.e. they have the same frequency.

Hence, they have the same pitch. (1A)

(b) Notes X, Y and Z have the same

loudness. (1A)

It is because the amplitudes of their

waveforms are the same. (1A)

(c) Each note given out by an instrument

has overtones added to the fundamental

note. (1A)

For notes of the same frequency given

out by different instruments, they have

the same fundamental note but different

overtones, (1A)

the frequencies of which are multiples

of the frequency of the fundamental

note and the amplitudes of which are

different. (1A)

Therefore, the resultant sound wave

(overtones added to the fundamental

note) given by different instruments

sounds differently.

(d) If the note has a sound intensity level of

140 dB, it causes permanent damage to

the ear. (1A)

10 (a) Both sound and light are waves (1A)

and show reflection, refraction,

diffraction and interference. (1A)

Sound waves are longitudinal waves

while light waves are transverse waves.

(1A)

The travelling speed of sound waves in

the air is 330 m s–1 while the travelling

speed of light waves in the air is

3 108 m s–1. (1A)

(b)

For light waves:

(Decreased wavelength in water.) (1A)

(The wave bends to normal in water.)

(1A)

For sound waves:

(Decreased wavelength in water.) (1A)

(The wave bends to normal in water.)

(1A)

11 (a)



(Correct amplitude.) (1A)

(Correct period.) (1A)

(b) The waveform of Michael’s voice has a

larger amplitude. (1A)

The waveform of Michael’s voice

shows a quality different from Julia’s.

(1A)

(c)

(Same amplitude and period.) (1A)

(Correct quality.) (1A)

12 (a) Interference (1A)

(b) (i) Period

= time occupied by 10 divisions on

CRO / number of complete

waves in 10 divisions

=

= 2.22 103 s

f = (1M)

=

= 450 Hz (1A)

By v = f, (1M)

=

=

= 0.756 m (1A)


0.756 m.

(ii)

(Same frequency.) (1A)

(Smaller amplitude.) (1A)

(c) (i) Diffraction occurs when sound

waves pass through the gap and

(1A)

sound waves spread out after

passing through the gap. (1A)



(ii) Increasing the pitch of the sound

means decreasing the wavelength.

(1A)

Sound waves would spread at a

smaller degree after passing the

gap. (1A)

13 (a) (i) An ultrasound signal emitted by

the ship is reflected when it hits an

obstacle. (1A)

After receiving the echo, the

distance between the ship and the

obstacle can be found by

measuring the time elapsed from

emitting the signal to receiving the

echo. (1A)

(ii) Time taken for the signal to travel

from the ship to the sea

=

= 0.2 s

Depth of the sea bed

= speed time (1M)

= 1500 0.2

= 300 m (1A)

(iii) The wavelength of ultrasound is

smaller than that of audible sound.

Ultrasound diffracts less when it

passes around obstacles in the sea

and it can detect the locations of

obstacles more accurately. (1A)

(b) (i) X-ray is a kind of electromagnetic

wave which is produced by electric

and magnetic vibrations. (1A)

Ultrasound is a kind of sound that

we cannot hear and is produced by

mechanical vibrations. (1A)

X-ray is a transverse wave while

ultrasound is a longitudinal wave.

(1A)

The travelling speed of X-ray in a

vacuum (3 108 m s–1) is much

higher than the travelling speed of

ultrasound in air (340 m s–1). (1A)

(ii) The energy carried by X-ray is

much higher than that by

ultrasound. (1A)

Foetuses may be hurt when X-rays

hit them. (1A)

14 (a) (i)

(Correct reflection of waves at

each window surface.) (2 1A)

(ii)

(Correct window position.) (1A)

(Correct reflection of sound

waves.) (1A)

(b) (i) Interference of sound. (1A)

(ii) Loud and soft sounds are heard

along PQ. (1A)



At some positions, sound waves

from the loudspeakers reinforce

each other and loud sounds are

formed. (1A)

At some other positions, sound

waves from the loudspeakers

cancel each other and soft sounds

are formed. (1A)

(iii) By v = f,

=

=

= 0.17 m (1A)

Let x be the shortest possible

distance between the student and

speaker B.

Path difference

= x – 4

=

= 0.17 (1M)

x = 4.085 m (1A)

The shortest possible distance

between the student and speaker B

is 4.085 m.

(iv) If the frequency of the sound is

increased, the wavelength of the

sound waves decreases. (1A)

The nodal lines that join the places

of destructive interference increase

in number and become closely

spaced. That means the points of

destructive interference become

closely spaced. (1A)

If the frequency is further

increased, the points of destructive

interference become so close

together (1A)

that they are hardly aware. (1A)

(v) No. (1A)

For a song, there are sounds of

many frequencies merging

together. (1A)

Therefore, the sounds from the

loudspeakers are no longer

coherent. (1A)

(For effective communication.)

(1C)


16 (a) Ultrasound is very high frequency

sound. It is beyond human hearing /

humans cannot hear this frequency.

(2A)

(b) (i) By v = f, (1M)

= (1M)

=

= 0.03 m (1A)

The wavelength of the ultrasound

is 0.03 m.

Correct unit: m (1A)

(ii) Total distance = 2400 m (1A)

t = (1M)

=

= 1.6 s (1A)

The time it would take for the

ultrasound wave to travel from the



transmitter to the sea-bed and back

to the receiver is 1.6 s.

Correct unit: s (1A)

(c) A suggestion to include:

Ordinary sound spreads more;

The concentration of ordinary sound is

less / Ordinary sound is less intense /

The amplitude of ordinary sound

decreases rapidly;

The range of ordinary sound is limited /

Ordinary sound cannot travel far.

(3 1A)

17 (a) The time interval

= 1.4 5.0 10–6

= 7.0 10–6 s (1A)

The time interval during which the

ultrasound travels in the organ is

7.0 10–6 s.

(b) s = vt

= 1.6 103 7.0 10–6

= 0.0112 m (1A)

The distance travelled by ultrasound

through the organ is 0.0112 m.



Physics in articles (p. 225) 1 (a) The barrier works by reflecting the

sound waves back into the roads. (1A)

(b) An effective barrier should have a solid,

continuous surface without any

openings or holes. (1A)

Otherwise, sound waves can pass

through the openings with the process of

diffraction. (1A)

An effective barrier must be long and

tall enough (1A)

so as to create a significant acoustical

shadow. (1A)

(c) No. (1A)

Since noise of low frequency has longer

wavelength, it bends more when passing

the barrier. Hence, flats in the shadow

of the barrier are still affected by the

noise of low frequency and barrier is not

effective to such noise. (1A)

2 (a) When we speak, vocal cords come

together and air exhaled from the lung

passes through the larynx, (1A)

setting the vocal cords to vibrate and

produce sound. (1A)

(b) One controls the tension of the vocal

cords for different pitches of sound.

(1A)

The larger the tension in the vocal cord,

the higher is the pitch of the sound. (1A)

One controls the flow of air in the

larynx for different loudness of sound.

(1A)

The larger the flow of air in larynx, the

higher is the loudness of sound. (1A)

(c) By v = f,

= = = 4.25 m (1A)

The longest wavelength of the sound

that is spoken by a man is 4.25 m.

By v = f,

= = = 1.7 m (1A)

The shortest wavelength of the sound

that is spoken by a man is 1.7 m.

By v = f,

= = = 2.27 m (1A)

The longest wavelength of the sound

that is spoken by a woman is 2.27 m.

By v = f,

= = = 0.85 m (1A)

The shortest wavelength of the sound

that is spoken by a woman is 0.85 m.

(d) Since vocal cords of boys will be

thickened during puberty (1A)

and thick vocal cords do not have as

large tension as thin vocal cords, (1A)

men usually have a deeper voice than

woman.


Date post:	14-Nov-2014
Category:	Documents
Upload:	api-3826695
View:	539 times
Download:	2 times

Chapter 13

Documents