Sound Waves - NCERT Solutions Class 9 Science Exploration

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Sound Waves: Characteristics and Applications – NCERT Solutions Class 9 Science Exploration includes all the questions with solutions given in NCERT Class 9 Science Exploration textbook.

NCERT Solutions Class 9

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Sound Waves – NCERT Solutions

Q.1: Two astronauts are repairing the arm of a space station together during a spacewalk. Can they talk to each other and hear the sounds of metal clanking as they do on the Earth?

Solution: No, the astronauts cannot talk to each other directly in space or hear the sound of metal clanking.

Sound needs a material medium (like air, water, or solids) to travel. In space, there is no atmosphere (vacuum), so sound waves cannot propagate. Therefore, even though they may speak or the metal may collide, the sound cannot reach the other astronaut.

Q.2: How do most bats use sound to locate their prey in the dark at night?

Solution: Bats use a method called echolocation to locate their prey in the dark.

They produce ultrasonic sound waves (high-frequency sounds not audible to humans). These sound waves travel, hit objects (like insects), and reflect back as echoes. By detecting the reflected sound, bats can determine the location, distance, and even the size of their prey.

Q.3: Explore various ways of producing sound.

Solution: Sound is produced when an object vibrates. There are several ways of producing sound:

Plucking: Vibrations are produced when strings are plucked, e.g., guitar, sitar.
Striking: Sound is produced when objects are struck, e.g., drum, bell.
Blowing: Vibrations of air column produce sound, e.g., flute, whistle.
Friction (rubbing): Sound is produced due to rubbing of surfaces, e.g., violin bow on strings.

Thus, in all cases, sound is produced due to vibrations of objects or air columns.

Q.4: Make a list of different types of musical instruments and identify their vibrating parts which produce sound.

Solution:

Type of musical instrument	Examples	Vibrating part that produces sound
String instruments	Guitar, Sitar, Violin	Stretched strings
Wind instruments	Flute, Trumpet, Clarinet	Air column inside the instrument
Membrane instruments	Drum, Tabla	Stretched membrane (skin)
Solid instruments (Percussion)	Bell, Cymbals	Whole body of the instrument
Keyboard instruments	Piano	Stretched strings inside (struck by hammers)

Q.5: Assertion (A): We cannot hear the sound of a bell ringing in a closed jar after most of the air is pumped out.
Reason (R): Sound requires a medium to travel.

Options:
(1) Both A and R are true and R is the correct explanation of A. ✅
(2) Both A and R are true but R is not the correct explanation of A.
(3) A is true but R is false.
(4) A is false but R is true.

Explanation: Sound cannot travel in vacuum. When air is removed from the jar, there is no medium for sound to propagate, so the sound of the bell cannot be heard.

Q.6: Assertion (A): Compressions and rarefactions move through the medium.
Reason (R): Individual particles of the medium continuously move forward with the wave.

Options:
(1) Both A and R are true and R is the correct explanation of A.
(2) Both A and R are true but R is not the correct explanation of A.
(3) A is true but R is false. ✅
(4) A is false but R is true.

Explanation: Assertion (A): True
Compressions and rarefactions move through the medium as a sound wave propagates.

Reason (R): False
The particles of the medium do not move forward with the wave; they only oscillate back and forth about their mean position.

Q.7: When sound travels from a tuning fork to your ear, which of the following actually reaches your ear?

Options:
(1) Air particles near the tuning fork
(2) Energy carried by sound waves ✅
(3) The tuning fork material
(4) A continuous stream of compressed air

Explanation: Sound travels through a medium as a wave. The particles of air only vibrate back and forth and do not move from the source to the ear. What is actually transferred is energy, not matter.

Q.8: If the frequency of a sound wave produced by an oscillating piston of a long tube filled with air is 20 Hz, then how many oscillations does the piston complete per minute?

Solution: Given:
Frequency of sound wave = 20 Hz

Step 1: Meaning of frequency

20 Hz means the piston completes 20 oscillations per second.

Step 2: Oscillations in 1 minute

1 minute = 60 seconds

Total oscillations in 1 minute = 20 {tex}\times{/tex} 60 = 1200
The piston completes 1200 oscillations per minute.

Q.9: For the sound wave represented by the graph shown in the figure, what is half of its wavelength?

Solution: From the graph:

One complete wavelength is the distance between two consecutive compressions (or rarefactions).
From the figure, this distance is 3.0 cm.

So,

{tex} \lambda=3.0 {~cm}{/tex}

Half of wavelength:

{tex} \frac{\lambda}{2}=\frac{3.0}{2}=1.5 {~cm}{/tex}

Q.10: Table shows the speed of sound in a few media at atmospheric pressure.

State	Substance/Medium	Approximate speed
Solid	Steel	{tex}5000 {~m} {s}^{-1}{/tex}
Liquid	Water	{tex}1500 {~m} {s}^{-1}{/tex}
Gas	Air	{tex}340 {~m} {s}^{-1}{/tex}

Compare the speeds in different media by finding the ratio of

the speed of sound in water with respect to the speed in the air.
the speed of sound in steel with respect to the speed in the water.

Solution: Given:

Speed in water {tex}=1500 {~m} / {s}{/tex}
Speed in air {tex}=340 {~m} / {s}{/tex}
Speed in steel {tex}=5000 {~m} / {s}{/tex}

Ratio of speed in water to speed in air
{tex} \frac{1500}{340}=\frac{150}{34}=\frac{75}{17} \approx 4.41{/tex}
{tex} 75: 17 \text { (or approximately } 4.41: 1 \text { ) }{/tex}
Ratio of speed in steel to speed in water {tex} \frac{5000}{1500}=\frac{50}{15}=\frac{10}{3} \approx 3.33{/tex} {tex} 10: 3 \text { (or approximately } 3.33: 1 \text {)}{/tex}

Q.11: An experiment is being set up that requires echoes to arrive at least 0.2 s after the emission of sound. What minimum distance should a reflecting surface be placed at? Assume the speed of sound to be 343 m s^-1.

Solution: For an echo, the sound has to travel to the reflecting surface and back, so the total distance covered is twice the distance to the surface.

Given:

Speed of sound {tex}v=343 {~m} / {s}{/tex}
Time delay {tex}t=0.2 {~s}{/tex}

Step:
Total distance traveled by sound:

{tex} \text {Total distance }=v \times t=343 \times 0.2=68.6 {~m}{/tex}
Since this is a round trip, the one-way distance (actual distance to the reflecting surface) is:

{tex} \frac{68.6}{2}=34.3 {~m}{/tex}

Q.12: Sound travels much farther in water than light, and thus, is used for various underwater applications. A sonar signal sent to find the depth of ocean takes 4 s to return. What is the depth of the ocean at that location if the speed of sound in seawater is 1500 m s^-1?

Solution: Given:

Speed of sound in seawater {tex}v=1500 {~m} / {s}{/tex}
Total time {tex}t=4 {~s}{/tex}

Step:
Total distance traveled by the sound:

{tex} \text {Distance}=v \times t=1500 \times 4=6000 {~m}{/tex}
Since this is a round trip, the actual depth is half of this:

{tex} \frac{6000}{2}=3000 {~m}{/tex}

Q.13: If there are 10 density oscillations in 2 seconds at a given position, then calculate the (i) frequency of sound wave, and (ii) its time period.

Solution: Given:
Number of oscillations = 10
Time = 2 s

Frequency of sound wave
We know that frequency is the number of oscillations per second.
{tex}\text { Frequency }({f}) =\frac{\text { Number of oscillations }}{\text { Time taken }}{/tex}
{tex}f =\frac{10}{2}=5 {~Hz}{/tex}
Time period of sound wave
Time period is the time taken for one oscillation.
{tex}T=\frac{1}{f}{/tex}
{tex}T=\frac{1}{5}=0.2 {~s}{/tex}

Q.14: Human hearing roughly spans 20 Hz to 20 kHz. What are the corresponding wavelengths in air for these two frequencies? Use the speed of sound in air as 344 m s^-1.

Solution: Using the relation between wavelength {tex}(\lambda){/tex}, frequency {tex}(\nu){/tex}, and speed ({tex}v{/tex}),
speed of the wave {tex}={/tex} frequency × wavelength
Therefore, wavelength {tex}\lambda=\frac{\text { speed of the wave }}{\text { frequency }}{/tex}

For {tex}\nu =20 {~Hz}, \lambda=\frac{344 {~m} {~s}^{-1}}{20 {~s}^{-1}}=17.2 {~m}{/tex}
For {tex}\nu =20 {kHz}=20000 {~Hz}, {/tex} {tex}\lambda=\frac{344 {~m} {~s}^{-1}}{20000 {~s}^{-1}}=0.0172 {~m}=1.72 {~cm}{/tex}

The wavelength of sound in air corresponding to the frequency (i) 20 Hz is 17.2 m, and (ii) 20000 Hz is 1.72 cm.

Q.15: During a thunderstorm, lightning is seen before thunder is heard because sound travels much slower than light. If the time delay between seeing the lightning flash and hearing the thunder is measured to be 5 s, estimate the distance to the lightning strike. Use the speed of sound in air as {tex}340 {m} {s}^{-1}{/tex}. Assume that light (speed {tex}=300000 {~km} {s}^{-1}{/tex}) reaches you almost instantaneously.

Solution: Distance {tex}=v \times t=340 {~m} {s}^{-1} \times 5 {~s}=1700 {~m}{/tex}
Lightning struck about 1.7 km away.

Q.16: From the graphical representation of a sound wave propagating in steel, find its wavelength. Calculate its frequency and time period if the speed of sound in steel is 5000 m s^-1.

Solution: From the graph, the wavelength {tex}\lambda=50 {~m}{/tex}
Using Eq. ({tex}v=\lambda \times \nu{/tex}), the frequency of the sound wave is

{tex} \nu=\frac{v}{\lambda}=\frac{5000 {~m} {~s}^{-1}}{50 {~m}}=100 {~Hz}{/tex}

Using Eq. ({tex}v=\frac{1}{T}{/tex}), the time period of the sound wave is

{tex} T=\frac{1}{\nu}=\frac{1}{100 {~Hz}}=0.01 {~s}{/tex}

Q.17: You clap in an empty corridor and hear an echo after 0.5 s. If the speed of sound in air is 340 m s^-1, calculate your distance from the wall.

Solution: Sound travels to the wall and back, thus,
distance from wall {tex}=\frac{v \times t}{2}=\frac{340 {~m} {s}^{-1} \times 0.5 {~s}}{2}=85 {~m}{/tex}

Q.18: A naval sonar signal sent into seawater returns after 0.90 s. The speed of sound in seawater is 1530 m s^-1. How far is the object?

Solution: Time taken for the signal to reach the object and travel back = 0.90 s
Time taken to reach the object is half of above time {tex}=\frac{0.90 {~s}}{2}=0.45 {~s}{/tex}
Thus, distance {tex}={/tex} speed × time {tex}=1530 {~m} {s}^{-1} \times 0.45 {~s}=688.5 {~m}{/tex}

Q.19: Which observation best supports the idea that sound is a mechanical wave?

Options:
(1) Sound shows reflection
(2) Sound needs a medium to propagate ✅
(3) Sound has frequency
(4) Sound carries energy

Explanation: Mechanical waves are those waves which require a material medium for propagation. Sound is a mechanical wave because it cannot travel in vacuum and needs particles of a medium (like air, water, or solids) to move and transfer energy.

Reflection, frequency, and energy are properties common to many types of waves (including light, which is not mechanical).
The need for a medium is the distinguishing feature of mechanical waves.

Q.20: For a sound wave propagating in a medium, increasing its frequency will increase its

Options:
(1) wavelength
(2) speed
(3) number of compressions per second ✅
(4) time period

Explanation: Frequency is defined as the number of oscillations (or compressions and rarefactions) per second.

If frequency increases → number of compressions per second increases
Wavelength decreases (inverse relation)
Speed remains constant in the same medium
Time period decreases {tex}\left(T=\frac{1}{f}\right){/tex}

Q.21: If 20 compressions pass a point in 4 seconds, the frequency is

Options:
(1) 80 Hz
(2) 5 Hz ✅
(3) 10 Hz
(4) 0.2 Hz

Explanation: Given:

Number of compressions {tex}=20{/tex}
Time {tex}=4{/tex} seconds

Concept:
Frequency is defined as the number of vibrations (or compressions) per second.

{tex} \text { Frequency }=\frac{\text { Number of compressions }}{\text { Time }}{/tex}
{tex} f=\frac{N}{t}{/tex}
Calculation:

{tex} f=\frac{20}{4}=5 {~Hz}{/tex}

Q.22: 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.

Solution: The human brain possesses a property called persistence of hearing. When we hear a sound, the sensation remains in our brain for approximately 0.1 seconds.

The Threshold for Echo: To hear a distinct echo, the reflected sound must reach our ears after the first sound has faded- meaning the time interval between the original sound and the reflected sound must be at least 0.1 s.
The Given Scenario: In this case, the reflected sound arrives in only 0.05 s.
The Result: Since 0.05 s is less than 0.1 s ({tex}0.05\text{ s} < 0.1\text{ s}{/tex}), the original sound and the reflected sound will overlap and “fused” together in the brain.

Q.23: Graphs representing two sound waves are given in the figure. 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?

Solution:

Greater Wavelength
Wave (a) has the greater wavelength.
Reasoning: Wavelength ({tex}\lambda{/tex}) is the horizontal distance between two consecutive peaks (crests) or two consecutive troughs. Looking at the X-axis (horizontal), the distance between the peaks in graph (a) is significantly larger than the distance between the peaks in graph (b). Since the scales are the same, a longer physical distance on the graph represents a greater wavelength.
Smaller Amplitude
Wave (a) has the smaller amplitude.
Reasoning: Amplitude is the maximum vertical displacement of the wave from its central equilibrium position (the X-axis). Looking at the Y-axis (vertical), the “height” of the peaks in wave (a) is much shorter compared to the peaks in wave (b). Therefore, wave (a) represents a softer sound with smaller amplitude, while wave (b) represents a louder sound with greater amplitude.

Q.24: The sound waves emitted by three sources A, B and C are represented in the figure. If the frequency of A is maximum and C is minimum, identify the corresponding curves, and mark A, B and C on them.

Solution: From the figure:

The wave with maximum number of oscillations (shortest wavelength) has maximum frequency → A
The wave with minimum oscillations (longest wavelength) has minimum frequency → C
The remaining wave is B

Q.25: Draw a graph to represent a sound wave for which the density amplitude is 3 units and wavelength is 4 cm.

Solution:

Q.26: 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?

Solution: In space, there is vacuum, so sound waves cannot propagate. Therefore, during a spacecraft explosion, only light should be seen, not heard.

Also, even if sound were possible, light travels much faster than sound, so they cannot be observed simultaneously.

Q.27: 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.

Solution: Given:
Wavelength {tex}(\lambda)=3.44 {~m}{/tex}
Wave speed {tex}(v)=344 {~m} / {s}{/tex}

Concept:

{tex} v=\lambda \nu \text { and } T=\frac{1}{\nu}{/tex}

Calculation:

{tex}\nu =\frac{v}{\lambda}=\frac{344}{3.44}=100 {~Hz}{/tex}
{tex}T =\frac{1}{\nu}=\frac{1}{100}=0.01 {~s}{/tex}

Q.28: 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?

Solution: Given:
Time for echo {tex}({t})=5 {~s}{/tex}
Speed of sound in seawater {tex}(v)=1525 {~m} / {s}{/tex}

Concept:
Sound travels to the object and back, so total distance {tex}=2 {~d}{/tex}

{tex} d=\frac{v \times t}{2}{/tex}

Calculation:

{tex} d=\frac{1525 \times 5}{2}=\frac{7625}{2}=3812.5 {~m}{/tex}

Q.29: 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.

Solution: Given:
Distance from obstacle (d) = 1.2 m
Speed of sound {tex}(v)=345 {~m} / {s}{/tex}

Concept:
The ultrasonic wave travels to the obstacle and back, so total distance {tex}=2 {~d}{/tex}

{tex} t=\frac{2 d}{v}{/tex}

Calculation:

{tex}t=\frac{2 \times 1.2}{345}=\frac{2.4}{345}{/tex}
{tex}t \approx 0.007 {~s}{/tex}

Q.30: The speed of sound in air is about {tex}331 {~m} {s}^{-1}{/tex} at {tex}0^{\circ} {C}{/tex} and nearly {tex}344 {~m} {s}^{-1}{/tex} at {tex}22^{\circ} {C}{/tex}. Roughly how much extra time will the sound of thunder take to travel a distance of 1720 m, if the air temperature changes from {tex}22^{\circ} {C}{/tex} to {tex}0^{\circ} {C}{/tex}? Assume that all other conditions remain unchanged.

Solution:

Given:
Distance (d) = 1720 m
Speed at {tex}22^{\circ} {C}\left({v}_1\right)=344 {~m} / {s}{/tex}
Speed at {tex}0^{\circ} {C}\left(v_2\right)=331 {~m} / {s}{/tex}

Concept: {tex} t=\frac{d}{v}{/tex}

Calculation:
Time at {tex}22^{\circ} {C}{/tex}:

{tex} t_1=\frac{1720}{344}=5 {~s}{/tex}

Time at {tex}0^{\circ} {C}{/tex}:

{tex} t_2=\frac{1720}{331} \approx 5.19 {~s}{/tex}

Extra time taken:

{tex} \Delta t=t_2-t_1=5.19-5=0.19 {~s}{/tex}

Q.31: The variation of density of medium for a sound wave propagating with a speed of 340 m s^-1 is shown in figure. Calculate the wavelength and frequency of the sound wave.

Solution: From the given density-distance graph of a sound wave:
The distance between two consecutive compressions (or rarefactions) = one wavelength ({tex}\lambda{/tex}).

From the figure, this distance is {tex}{4 0 ~ c m}={0 . 4 0 ~ m}{/tex}.

So,
wavelength {tex}\lambda=0.40 {~m}{/tex}

Now using the wave relation from NCERT:

{tex} v=f \lambda{/tex}

So,

{tex}f=v / \lambda{/tex}
{tex}f=340 / 0.40{/tex}
{tex}f=850 {~Hz}{/tex}

Q.32: The graphical representation of two sound waves A and B propagating at the same speed of 345 m s^-1 is shown in the figure. What is the wavelength of each of them? Also, calculate their frequencies.

Solution: Given:
Speed of sound {tex}(v)=345 {~m} / {s}{/tex}

Concept:
Wavelength is the distance between two consecutive crests.

{tex} v=\lambda f{/tex}
From the graph:

Wave A has wavelength {tex}2 {~cm}(0.02 {~m}){/tex}
Wave {tex}B{/tex} has wavelength {tex}4 {~cm}(0.04 {~m}){/tex}

For A:

{tex} f_A=\frac{345}{0.02}=17250 {~Hz}{/tex}

For B:

{tex} f_B=\frac{345}{0.04}=8625 {~Hz}{/tex}

Q.33: Two identical sound sources are placed at A and B - one in air and one submerged in water. 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 that of B, what is the ratio between the speeds of sound in air and water?

Solution: Given: Time taken by sound in air to return to {tex}{A}\left(t_A\right)\ is\ 4.5{/tex} times that of {tex}{B}\left(t_B\right){/tex} i.e., {tex}t_A=4.5 t_B{/tex}
To Find: Ratio of speeds of sound in air and water {tex}\left(\frac{v_{\text {air }}}{v_{\text {water }}}\right){/tex}
Concept: For the same distance, time taken is inversely proportional to speed.

{tex}\text {Time taken} \ t=\frac{\text { distance }}{\text { speed }}{/tex}

Let the distance between the source and the cliff be {tex}d{/tex}.
For sound in air: {tex}t_A=\frac{2 d}{v_{\text {air }}}{/tex}
For sound in water: {tex}t_B=\frac{2 d}{v_{\text {water }}}{/tex}
Given: {tex}t_A=4.5 t_B{/tex}
So, {tex}\frac{2 d}{v_{\text {air }}}=4.5 \times\left(\frac{2 d}{v_{\text {water }}}\right){/tex}

{tex} \frac{1}{v_{\text {air }}}=\frac{4.5}{v_{\text {water }}}{/tex}

{tex}\frac{v_{\text {water }}}{v_{\text {air }}} =4.5{/tex}
{tex}\frac{v_{\text {air }}}{v_{\text {water }}} =\frac{1}{4.5}=\frac{2}{9}{/tex}

The ratio between the speeds of sound in air and water is {tex}2: 9{/tex}.