MCE Cambridge IGCSETM Physics Full Solutions to Questions in Student’s Book
Chapter 14
Chapter 14
Sound
All questions (including those that are exam-style) and answers have been written by the authors.
In examinations, the way marks are awarded may be different. References to assessment and/or
assessment preparation are the publisher’s interpretation of the syllabus requirements and may
not fully reflect the approach of Cambridge Assessment International Education.
Chapter Opener [Page 223]
• In a battle scene on Earth, we can expect to hear the sounds of explosions, shootings, vehicles [e.g.
tanks, jet planes], people screaming, etc. We can hear these sounds because there are air, solids
and liquids around through which sound can travel to reach our ears.
• The misconception is that we can hear sound in outer space just like on Earth.
• Space is a vacuum. We cannot hear sound in space because sound requires a medium for its
transmission.
Enrichment (Activity) [Page 224]
No vibrations can be felt when making a hissing sound whereas vibrations can be felt on the vocal cord
when making a buzzing sound.
Quick Check [Page 225]
False
There are more air particles in a region of compression than in a region of rarefaction.
Let’s Practise 14.1 [Page 226]
1.
2.
(a) A longitudinal wave is a wave whereby the direction of vibration of the particles (such as air
molecules) is parallel to the direction in which the wave travels.
(b) Compressions are regions where air pressure is higher than the surrounding air pressure.
Rarefactions are regions where air pressure is lower than the surrounding air pressure.
No. The audible range for the human ear is 20 Hz to 20 000 hz.
Quick Check [Page 227]
False
Sound requires a medium for its transmission.
Let’s Practise 14.2 [Page 228]
1.
2.
No, as sound waves cannot pass through a vacuum between the spaceships.
Speed of sound =
!"#$%&'( $*%+(,,(!
$"-( $%.(&
=
/000 1#
333 m/s
Enrichment (Think) [Page 231]
The bat has very large ears. This suggests that it relies on sound for finding its food. The bat is likely to
use echolocation to catch its prey.
An effective way to catch this bat is to use a large net. Sound can pass through the spaces in-between
the net. The bat will not be able to detect the net. It will fly towards the net and get trapped.
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MCE Cambridge IGCSETM Physics Full Solutions to Questions in Student’s Book
Chapter 14
Let’s Practise 14.3 [Page 231]
1.
2.
3.
The multitude of echoes is caused by the multiple reflections of sound from the many reflecting
surfaces, such as the ceiling and the walls of the hall.
The pulse of sound travelled towards the seabed and reflected back in 1 second, this means that
the distance travelled is 2d, where d is the depth of the sea.
23
Speed of sound v =
4
54 /600 × /
Therefore, d =
=
= 750 m
2
2
Ultrasound is less hazardous than X-rays due to its lower energy.
Quick Check [Page 232]
True
The higher the frequency of the sound waves, the higher the pitch if the sound.
Enrichment (Activity) [Page 233]
The amount of air space in each bottle is different. The air molecules vibrate faster when the air space is
smaller. This will increase the pitch of the note produced. Several bottles can be used to make different
notes to play a song.
Let’s Practise 14.4 [Page 233]
1.
(a)
(b)
Amplitude
Frequency
2.
Loudness: Bullfrog louder than mosquito
Pitch: Mosquito higher pitch than bullfrog
Let’s Review [Pages 235–236]
Section A: Multiple-choice Questions
1.
C
Sound cannot travel through space, as it is a vacuum.
2.
C
3.
B
Boy B hears two claps: the first clap arrives directly from boy A, and the second clap is due to the
reflection (i.e. the echo) of boy A’s clap by the tall building. The direct clap travels 200 m to boy B,
while the echo travels 880 m (340 m + 340 m + 200 m = 880 m). The extra distance travelled by
the echo causes it to be received by boy B 2 s later than the direct clap. Hence, the speed of
880 m – 200 m
sound is
= 340 m/s
2s
4.
D
Section B: Short-answer and Structured Questions
1.
(a) When the hammer strikes the bell, the bell vibrates [1] and thus causing the air particles to
vibrate and we hear sound. [1]
(b) The vibrating bell causes layers of air particles around it to shift. This causes a series of
alternate high-pressure regions (compressions) and low-pressure regions (rarefactions) [1] to
travel outwards and propagate through air to reach the ear of the person. [1]
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14.2
MCE Cambridge IGCSETM Physics Full Solutions to Questions in Student’s Book
Chapter 14
2.
(a) Sound waves need a medium to travel in. [1] The vibrations of the ringing bell cause air
particles in the bell jar to vibrate. The air in the bell jar thus provides the medium to transmit the
sound waves from the ringing bell to the bell jar [1]. The sound waves are then transmitted from
the bell jar through the surrounding air to our ears [1]. The removal of the air reduces the
number of air particles in the bell jar [1]. Thus, the sound waves are transmitted from the bell to
the bell jar with decreasing effectiveness and the loudness of the bell decreases [1]. The faint
sound indicates that the bell jar still contains some air. This is because the vacuum pump does
not create a complete vacuum inside the bell jar.
(b) There will not be a significant reduction of the loudness [1] of the bell even when the air is
removed by the vacuum pump. Since the electric bell is in direct contact with the base, the
sound waves are propagated directly through the base [1] to the air surrounding the bell jar.
3.
(a)
Average time taken =
/.6/ 9 /.66 9 /.60
= 1.52 s [1]
1
Average speed of sound in air [1] =
!"#$%&'( 600
=
= 329 m/s [1]
$"-(
/.62
(b) The human reaction is not fast enough. A distance of 100 m is too close [1] and would not
allow observer B to react promptly to start and stop the stopwatch [1]. It is even possible that
he would see the flash and hear the sound simultaneously. [1]
4.
(a) Dogs can detect sound of frequencies above 20 000 Hz. [1]
(b) A transmitter sends ultrasound pulses into the body. The time taken for the ultrasound pulses
to be reflected from the surface of the internal body part, and received, is measured. [1] The
depth of the reflecting surface and the intensity of the reflected wave within the body can then
be derived to form an image of the internal body part. [1]
5.
(a) An echo is formed when a sound is reflected off hard, flat and large surfaces. [1]
(b) (i)
At A:
Time taken for sound to travel to and back from seabed = 0.4 s
Speed of sound in water = 1500 m/s
Let the depth of seabed at A be dA
23!
[1]
4
54 /600 × 0.:
dA =
=
= 300 m [1]
2
2
At B:
Time taken for sound to travel to and back from seabed = 0.4 s
Speed of sound in water = 1500 m/s
Let the depth of seabed at B be dB
v=
dB =
54 /600 × 0.:
=
= 300 m [1]
2
2
At C:
Time taken for sound to travel to and back from seabed = 0.6 s
Speed of sound in water = 1500 m/s
Let the depth of seabed at C be dC
dC =
54 /600 × 0.;
=
= 450 m [1]
2
2
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14.3
MCE Cambridge IGCSETM Physics Full Solutions to Questions in Student’s Book
Chapter 14
At D:
Time taken for sound to travel to and back from seabed = 0.6 s
Speed of sound in water = 1500 m/s
Let the depth of seabed at D be dD
dD =
54 /600 × 0.;
=
= 450 m [1]
2
2
At E:
Time taken for sound to travel to and back from seabed = 0.2 s
Speed of sound in water = 1500 m/s
Let the depth of seabed at E be dE
dE =
54 /600 × 0.2
=
= 150 m [1]
2
2
At F:
Time taken for sound to travel to and back from seabed = 0.2 s
Speed of sound in water = 1500 m/s
Let the depth of seabed at F be dF
dF =
54 /600 × 0.2
=
= 150 m [1]
2
2
(ii)
[1]
[1]
[1]
(iii)
6.
t=
23
2 × ;0
[1] =
= 0.08
0.8 ssec
[1]
5
/600
(a) (i) Frequency [1]
(ii) Amplitude [1]
(b) He will not succeed. [1] Only the amplitude of the note will be affected, but the frequency
remains the same. He will only succeed in producing louder notes. [2]
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14.4