Sound Waves: Characteristics and Application - Chapter 10 Exploration

Question 1
Which observation best supports the idea that sound is a mechanical wave? (i) Sound shows reflection (ii) Sound needs a medium to propagate (iii) Sound has frequency (iv) Sound carries energy
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Answer: (ii)
(ii) correct
— needing a medium is the defining feature of a mechanical wave.
Reflection, frequency and carrying energy are true of sound but are shared by many waves, so they do not specifically prove it is mechanical.
Option
Verdict & why
(i)
Sound shows reflection
Wrong
— light also reflects, and light is
not
a mechanical wave.
(ii)
Sound needs a medium to propagate
Correct
— needing a material medium is the
defining
test of a mechanical wave.
(iii)
Sound has frequency
Wrong
— every wave, mechanical or not, has a frequency.
(iv)
Sound carries energy
Wrong
— all waves carry energy, including light.
Back to: 10.3 Sound Waves

Question 2
For a sound wave propagating in a medium, increasing its frequency will increase its (i) wavelength (ii) speed (iii) number of compressions per second (iv) time period
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Answer: (iii)
Higher frequency = more oscillations per second =
more compressions per second
.
Speed depends on the medium (unchanged); wavelength and time period
decrease
as frequency rises.
Option
Effect of raising frequency
(i)
wavelength
Decreases
—
v=λν
with
v
fixed, so
λ∝ 1/ν
.
(ii)
speed
Unchanged
— speed depends only on the medium, not the source.
(iii)
number of compressions per second
Increases
— this
is
the frequency.
Correct answer.
(iv)
time period
Decreases
—
T=1/ν
, so a higher frequency means a shorter period.
Back to: 10.6.1 Wavelength, frequency and time period

Question 3
If 20 compressions pass a point in 4 seconds, the frequency is (i) 80 Hz (ii) 5 Hz (iii) 10 Hz (iv) 0.2 Hz
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Answer: (ii) 5 Hz
Frequency
=
compressions
time
=
20
4
=5
Hz
.
Option
Verdict & why
(ν=
20
4
)
(i)
80 Hz
Wrong
— that is
20× 4
, not
20÷ 4
.
(ii)
5 Hz
Correct
—
ν=
20
4
=5
oscillations per second.
(iii)
10 Hz
Wrong
— that would need 40 compressions in 4 seconds.
(iv)
0.2 Hz
Wrong
— that is
4÷ 20
, the time period, not the frequency.
Back to: 10.6.1 Wavelength, frequency and time period

Question 4
In a room, the reflected sound reaches the ear 0.05 s after its production. Will it produce an echo or reverberation? Justify your answer.
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The gap is
0.05 s
, which is less than 0.1 s, so the reflected sound is
not
heard as a separate echo.
A gap under 0.1 s (here 0.05 s) causes
reverberation
— the reflections overlap and the sound persists.
Echo or Reverberation
The gap is 0.05 s
Less than 0.1 s, so not a
separate echo
It is reverberation
Back to: 10.7.2 Reverberation

Question 5
Graphs representing two sound waves are given in Fig. 10.30. If the scales on the X and Y axes of the two graphs are the same, which of the two sound waves has (i) greater wavelength, and (ii) smaller amplitude?
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(i) The wave with crests spaced
farther apart
has the
greater wavelength
.
(ii) The wave whose curve rises and falls
less
from the average line has the
smaller amplitude
.
Reading Two Graphs
Wider crest spacing means
greater wavelength
Smaller rise from average
means smaller amplitude
Compare the two curves
Back to: 10.6.1 Wavelength, frequency and time period

Question 6
The sound waves emitted by three sources A, B and C are represented in Fig. 10.31. If the frequency of A is maximum and C is minimum, identify the corresponding curves, and mark A, B and C on them.
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Highest frequency (A) = curve with crests
closest together
(shortest wavelength).
Lowest frequency (C) = curve with crests
farthest apart
(longest wavelength); B is in between.
Marking A, B and C
A: highest frequency, crests closest C: lowest frequency, crests farthest B lies in between
Back to: 10.6.1 Wavelength, frequency and time period

Question 7
Draw a graph to represent a sound wave for which the density amplitude is 3 units and wavelength is 4 cm.
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Draw a density-distance curve whose height above and below the average line (the amplitude) is
3 units
.
Make the distance between two consecutive crests (the wavelength) equal to
4 cm
.
Drawing the Wave
Amplitude 3 units above
and below average
Wavelength 4 cm between
crests
Draw the density-distance
curve
Back to: 10.5 Graphical Representation of a Sound Wave

Question 8
In a movie, while showing the explosion of a spacecraft in space, a flash of light is shown along with sound at the same time. What are the errors in this depiction?
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Error 1:
space is a near vacuum, so the explosion
should make no sound
at all.
Error 2:
even with a medium, light travels far faster than sound, so the
flash and sound cannot arrive together
.
Errors in the Movie
Space is a vacuum, so no
sound
Light is far faster than
sound
Flash and sound cannot
arrive together
Back to: 10.2.1 Sound needs a medium to propagate

Question 9
A source produces a sound wave of wavelength 3.44 m. If the wave travels with a speed of 344 m s⁻¹ find its time period.
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v=λ×ν⇒ ν=
v
λ
=
344
3.44
=100
Hz
.
Time period
T=
1
ν
=
1
100
=
0.01
s
.
Time Period
nu = v / lambda = 344 / 3.44
= 100 Hz
T = 1 / nu
T = 0.01 s
Back to: 10.6.3 Speed of Sound

Question 10
A ship searching for a sunken ship sent a sonar signal and detected an echo after 5 s. If ultrasonic wave travels at 1525 m s⁻¹ in seawater, approximately how far down in the ocean is the wreckage of the sunken ship located?
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Time to reach the wreck
=
5
2
=2.5
s
.
Depth
=v× t=1525× 2.5=
3812.5
m
.
Sonar Depth
One-way time = 5 / 2 = 2.5 s
Depth = speed times time
1525 times 2.5 = 3812.5 m
Back to: 10.8 Ultrasonic and Infrasonic Waves, and their Applications

Question 11
A vehicle is fitted with an ultrasonic distance sensor as part of parking assistance system which provides echolocation, while the driver is reversing the vehicle. It emits ultrasonic wave (about 40 kHz) which is reflected by the obstacle. When the warning beep starts sounding at a distance of 1.2 m from the obstacle, how much time is taken by ultrasonic wave to travel to the obstacle and come back? Assume the speed of ultrasonic wave in air to be 345 m s⁻¹.
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Total path = to the obstacle and back
=2× 1.2=2.4
m
.
Time
=
distance
speed
=
2.4
345
≈
0.007
s
(about 7 ms).
Parking Sensor Time
Round trip = 2 times 1.2 =
2.4 m
Time = distance divided by
speed
2.4 / 345 is about 0.007 s
Back to: 10.8.1 Echolocation

Question 12
The speed of sound in air is about 331 m s⁻¹ at 0 °C and nearly 344 m s⁻¹ at 22 °C. Roughly how much extra time will the sound of thunder take to travel a distance of 1720 m, if the air temperature changes from 22 °C to 0 °C? Assume that all other conditions remain unchanged.
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Time at 22 °C
=
1720
344
=5
s
; time at 0 °C
=
1720
331
≈ 5.196
s
.
Extra time
≈ 5.196-5=
0.2
s
(roughly).
Extra Time for Thunder
At 344 m/s: 1720 / 344 = 5 s
At 331 m/s: about 5.196 s
Extra time is about 0.2 s
Back to: 10.6.3 Speed of Sound

Question 13
The variation of density of medium for a sound wave propagating with a speed of 340 m s⁻¹ is shown in Fig. 10.32. Calculate the wavelength and frequency of the sound wave.
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From Fig. 10.32 the wavelength (one full compression-to-compression) is
λ=8
cm
=0.08
m
.
Frequency
ν=
v
λ
=
340
0.08
=
4250
Hz
.
Wavelength and
Frequency
Wavelength = 8 cm = 0.08 m
nu = v / lambda = 340 / 0.08
Frequency = 4250 Hz
Back to: 10.6.1 Wavelength, frequency and time period

Question 14
The graphical representation of two sound waves A and B propagating at the same speed of 345 m s⁻¹ is shown in Fig. 10.33. What is the wavelength of each of them? Also, calculate their frequencies.
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Read each wavelength from Fig. 10.33 (crest-to-crest), then use
ν=
v
λ
with
v=345
m/s
.
The wave with the shorter wavelength (A) has the higher frequency; the longer one (B) the lower frequency.
Two Waves A and B
Read each wavelength from
the graph
Use nu = v / lambda with v
= 345
Shorter wavelength gives
higher frequency
Back to: 10.6.1 Wavelength, frequency and time period

Question 15
Two identical sound sources are placed at A and B — one in air and one submerged in water (Fig. 10.34). Both produce sounds at the same time, which travel horizontally to the vertical side of the cliff and come back. If the time taken by the sound to return to A is 4.5 times than that of B, what is the ratio between the speeds of sound in air and water?
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Same distance, so time is inversely proportional to speed:
t
A
t
B
=
v
B
v
A
=4.5
.
Hence
v
air
v
water
=
v
A
v
B
=
1
4.5
≈
1:4.5
(sound is faster in water).
Ratio of Speeds
Same distance, so time is
inverse of speed
t_A / t_B = v_B / v_A = 4.5
v_air to v_water is about 1
to 4.5
Back to: 10.6.3 Speed of Sound

Project 1
Many people use earphones extensively these days. Find out the research studies on the impact of excessive earphone use on hearing, how hearing is tested, and the decibel ranges for mild, moderate and severe hearing loss. Also find out the government schemes for aids, appliances and free cochlear implants, and write an article on your findings.
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Research audiograms and safe-listening limits (e.g. the 60/60 rule), summarise dB thresholds for hearing-loss categories, list relevant government schemes, and present it as a short article with sources.

Project 2
Make a cone using poster paper or cardboard and adhesive tape. Cover a mobile phone playing music with the cone and compare the loudness with and without it. Use an app to measure the sound in both cases, try different shapes, and record your observations.
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The cone channels and reinforces the sound in one direction, so the measured loudness is higher with the cone. Narrower or longer cones focus the sound differently — record the dB readings for each shape.

Project 3
How does the curved design of ceilings and walls behind the stage in concert and conference halls improve the quality of sound for the audience compared to flat surfaces? You may consult an architect or search the internet.
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Curved surfaces reflect and spread sound evenly toward the whole audience and reduce dead spots and harsh echoes, giving clearer sound than flat walls, which reflect sound unevenly.

Project 4
Carry out a simple activity to measure the speed of sound with a friend in a large open ground (200 m or more), using balloons and a stopwatch: time the gap between seeing a balloon burst and hearing it, then divide the measured distance by the average time.
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Because light reaches you almost instantly, the timed gap is the sound’s travel time. Dividing distance by the average time gives a speed close to 346 m/s at 25 °C. Repeat and average to reduce error.

Project 5
Explore internet resources to study the effect of humidity and temperature on the speed of sound, for example the PhET sound-waves simulation, Chrome Music Lab, and the Phyphox experiments.
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Use the simulations to vary temperature and humidity and note that the speed of sound rises as either increases; summarise the trend with a short table or graph of your observations.
🔑 The Quest Continues…
Sound helps us explore places beyond human hearing.
Space probes have recorded the first sounds from Mars.
Scientists time distant earthquakes to track ocean temperature.
Biologists use the buzz of mosquitoes to identify disease carriers.
Researchers listen to soil microbes to study soil biodiversity.
Keep listening, and keep questioning!