---
title: "NCERT Solutions Class 9 Science Sound Waves: Characteristics and Applications"
url: https://www.swavid.com/science/class/9/chapter/sound-waves-characteristics-and-applications/ncert-solutions
dateModified: 2026-10-07T16:38:31+00:00
---

# NCERT Solutions Class 9 Science Sound Waves: Characteristics and Applications

This chapter contains various questions designed to test a student's understanding of sound waves, their properties, and their applications. The exercises cover topics ranging from the production and propagation of sound to its graphical representation, human perception, and reflection.

Free PDF (15 pages): https://www.swavid.com/api/seo/pdf/ncert/science/class-9/swavid-ncert-solutions-class-9-science-chapter-10-sound-waves-characteristics-and-applications-3476131613.pdf

## Think It Over

### Question 1

*2 marks · Very short answer*

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**

1. No, the astronauts cannot talk directly or hear the metal clanking as they do on Earth.
2. Outer space is a near vacuum with no material medium, and sound waves require a medium to propagate.

**Answer:** No, they cannot hear each other directly because sound needs a material medium to propagate, and outer space is a near vacuum.

> Common mistake: Thinking sound can travel through empty space just like light does.

### Question 2

*2 marks · Very short answer*

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

**Solution**

1. Bats emit short bursts of high-frequency ultrasonic waves as they fly.
2. These waves get reflected back from obstacles and prey as echoes, allowing bats to determine their position through echolocation.

**Answer:** Bats use echolocation by emitting short bursts of ultrasonic waves and sensing the echoes reflected from their prey and surroundings.

> Common mistake: Forgetting to mention that bats use ultrasonic waves and echoes.

## Pause and Ponder

### Question 1

*3 marks · Short answer*

Explore various ways of producing sound.

**Solution**

1. Sound is produced by the vibration of objects, which refers to the periodic to and fro motion of an object.
2. Common methods include plucking a stretched string, striking a metal object, or blowing air through a hollow pipe to vibrate the air column inside.
3. Other methods include striking membranes, such as in drums, or rubbing body parts together, as observed in some insects like grasshoppers and crickets.

**Answer:** Sound is produced through various methods of vibration, including plucking strings, striking metal objects or membranes, and vibrating air columns.

> Common mistake: Failing to mention that vibration is the fundamental cause of sound production in all these methods.

### Question 2

*3 marks · Short answer*

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

**Solution**

1. String instruments like the sitar, veena, and tanpura produce sound through the vibration of their stretched strings.
2. Wind instruments like the bansuri (flute) produce sound through the vibration of the air column inside the hollow pipe.
3. Percussion instruments like the tabla and mridangam produce sound through the vibration of their stretched membranes.

**Answer:** Musical instruments produce sound through vibrating strings (sitar), air columns (bansuri), or membranes (tabla).

> Common mistake: Confusing the vibrating part of a wind instrument with the instrument body itself instead of the air column.

## Pause and Ponder

### Question 3

*1 mark · Assertion and reason*

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.
Choose the correct statement:

- Both A and R are true, but R is not the correct explanation of A.
- Both A and R are true, and R is the correct explanation of A.
- A is true, but R is false.
- A is false, but R is true.

**Solution**

1. Sound requires a material medium to propagate, which is demonstrated by the vacuum bell jar experiment.
2. As air is pumped out, the sound becomes fainter and eventually cannot be heard when a near vacuum is reached.

**Answer:** Both A and R are true, and R is the correct explanation of A.

> Common mistake: Thinking sound can travel through vacuum just because light can.

## Pause and Ponder

### Question 4

*1 mark · Assertion and reason*

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

- Both A and R are true, but R is not the correct explanation of A.
- Both A and R are true, and R is the correct explanation of A.
- A is true, but R is false.
- A is false, but R is true.

**Solution**

1. In a sound wave, compressions and rarefactions travel through the medium.
2. However, the individual particles of the medium do not continuously move forward with the wave, but only oscillate back and forth about their mean positions.

**Answer:** (iii) A is true, but R is false.

> Common mistake: Students often confuse the propagation of the disturbance (wave) with the net motion of medium particles.

## Pause and Ponder

### Question 5

*1 mark · MCQ*

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

- Air particles near the tuning fork
- Energy carried by sound waves
- The tuning fork material
- A continuous stream of compressed air

**Solution**

1. During the propagation of a sound wave through a medium, particles of the medium only vibrate about their mean positions without flowing with the wave.
2. It is the disturbance and the energy carried by the sound waves that travel from the source to the ear, making option (ii) correct.

**Answer:** (ii) Energy carried by sound waves

> Common mistake: Students often think air particles travel all the way from the source to the ear.

## Pause and Ponder

### Question 6

*3 marks · Short answer*

The variation of density of the medium for two sound waves is shown in Fig. 10.17 (a) and (b). Label compression and rarefaction by C and R on it. In the graph given in Fig. 10.17 (c) and (d), label the axes and draw the curves corresponding to Fig. 10.17 (a) and (b).

**Part (i)**

1. Identify the regions of higher density in Fig. 10.17 (a) and (b) and label them as C for compression.
2. Identify the regions of lower density between the compressions and label them as R for rarefaction.

Answer (i): Compressions and rarefactions are labelled with C and R respectively on the density variation diagrams.

**Part (ii)**

1. Draw coordinate axes for graphs (c) and (d), labelling the horizontal x-axis as 'Distance' and the vertical y-axis as 'Density'.
2. Draw a horizontal dashed line representing the average density of the medium.
3. Draw smooth sinusoidal curves oscillating above and below the average density line, matching the density patterns shown in figures (a) and (b) respectively.

Answer (ii): The axes are labelled as 'Distance' (x-axis) and 'Density' (y-axis), and the corresponding wave curves are drawn for figures (c) and (d).

**Answer:** The compressions (regions of higher density above average) and rarefactions (regions of lower density below average) are labelled as C and R respectively. In the graphical representation, the x-axis represents distance and the y-axis represents density, with sinusoidal curves oscillating above and below the average density line.

> Common mistake: Labelling distance on the y-axis and density on the x-axis, or failing to mark the average density line.

## Pause and Ponder

### Question 7

*3 marks · Short answer*

Conduct Activity 10.1 once again with a thick rubber band and then with a thin rubber band. Does the thin rubber band vibrate faster than the thick rubber band? If yes, how do the frequency and time period of the sound produced by the thin rubber band differ from that of the thick rubber band?

**Solution**

1. Yes, the thin rubber band vibrates faster than the thick rubber band when plucked.
2. The sound produced by the thin rubber band has a higher frequency than that produced by the thick rubber band.
3. Since frequency and time period are inversely related ($T = \frac{1}{\nu}$), the thin rubber band has a smaller time period compared to the thick rubber band.

**Answer:** The thin rubber band vibrates faster, producing a sound with a higher frequency and a smaller time period than the thick rubber band.

> Common mistake: Confusing the relationship between frequency and time period.

### Question 8

*3 marks · Numerical*

If the frequency of a sound wave produced by an oscillating piston of a long tube filled with air is $20\text{ Hz}$, then how many oscillations does the piston complete per minute?

**Solution**

1. Given: Frequency $\nu = 20\text{ Hz}$, time $t = 1\text{ minute} = 60\text{ s}$
2. Formula: Number of oscillations = $\nu \times t$
3. Substitution: Number of oscillations = $20\text{ s}^{-1} \times 60\text{ s}$
4. Result: $1200$ oscillations

**Answer:** 1200 oscillations

> Common mistake: Forgetting to convert minutes into seconds.

### Question 9

*3 marks · Numerical*

For the sound wave represented by the graph shown in Fig. 10.19, what is half of its wavelength?

**Solution**

1. From the figure in the textbook (Fig. 10.19), the wavelength $\lambda$ of the sound wave (distance between two consecutive crests or troughs) is $3.0\text{ cm}$.
2. Formula: Half of the wavelength = $\frac{\lambda}{2}$
3. Substitution: Half of the wavelength = $\frac{3.0\text{ cm}}{2}$
4. Result: $1.5\text{ cm}$

**Answer:** $1.5\text{ cm}$

> Common mistake: Taking the full wavelength instead of half.

## Pause and Ponder

### Question 10

*3 marks · Numerical*

Table 10.1 shows the speed of sound in a few media at atmospheric pressure. Compare the speeds in different media by finding the ratio of (i) the speed of sound in water with respect to the speed in the air. (ii) the speed of sound in steel with respect to the speed in the water.

**Part (i)**

1. Given: Speed of sound in water ($v_{\text{water}}) = 1500~\text{m s}^{-1}$, speed of sound in air ($v_{\text{air}}) = 340~\text{m s}^{-1}$.
2. Ratio = $\frac{v_{\text{water}}}{v_{\text{air}}} = \frac{1500}{340} = 4.41$.

Answer (i): $4.41$

**Part (ii)**

1. Given: Speed of sound in steel ($v_{\text{steel}}) = 5000~\text{m s}^{-1}$, speed of sound in water ($v_{\text{water}}) = 1500~\text{m s}^{-1}$.
2. Ratio = $\frac{v_{\text{steel}}}{v_{\text{water}}} = \frac{5000}{1500} = 3.33$.

Answer (ii): $3.33$

**Answer:** The ratio of speeds in water to air is $4.41$, and in steel to water is $3.33$.

> Common mistake: Inverting the ratio by dividing the speed in air by the speed in water instead of with respect to air.

## Pause and Ponder

### Question 11

*3 marks · Numerical*

Two friends are standing along a steel fence at a distance of $340\text{ m}$ from each other (Fig. 10.23). Gunjan places her ear over the fence and her friend knocks the fence with a metal object. Using the values of the speed of sound in steel and air given in Table 10.1, calculate the time difference between the sound that reached Gunjan through the air and the steel. Would it have been possible for her to distinguish between the two sounds? (The time interval between two sounds must be at least $0.1\text{ s}$ to be heard separately.)

**Solution**

1. Given: Distance between the friends $d = 340\text{ m}$, speed of sound in steel $v_{\text{steel}} = 5000\text{ m s}^{-1}$, speed of sound in air $v_{\text{air}} = 340\text{ m s}^{-1}$.
2. Formula: Time taken $t = \frac{\text{distance}}{\text{speed}}$.
3. Time taken by sound through steel $t_{\text{steel}} = \frac{340\text{ m}}{5000\text{ m s}^{-1}} = 0.068\text{ s}$.
4. Time taken by sound through air $t_{\text{air}} = \frac{340\text{ m}}{340\text{ m s}^{-1}} = 1.0\text{ s}$.
5. Time difference $\Delta t = t_{\text{air}} - t_{\text{steel}} = 1.0\text{ s} - 0.068\text{ s} = 0.932\text{ s}$.
6. Since the time difference ($0.932\text{ s}$) is greater than $0.1\text{ s}$, it would be possible for her to distinguish between the two sounds.

**Answer:** Time difference is $0.932\text{ s}$; yes, she can distinguish between the two sounds.

> Common mistake: Dividing the wrong speeds or incorrect subtraction of the times.

## Pause and Ponder

### Question 12

*3 marks · Numerical*

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

**Solution**

1. Given: Speed of sound ($v$) = $343\text{ m s}^{-1}$, time ($t$) = $0.2\text{ s}$
2. Formula: Total distance travelled by sound = $\text{speed} \times \text{time}$
3. Substitution: $\text{Total distance} = 343\text{ m s}^{-1} \times 0.2\text{ s} = 68.6\text{ m}$
4. Result: Minimum distance of reflecting surface = $\frac{68.6}{2} = 34.3\text{ m}$

**Answer:** $34.3\text{ m}$

> Common mistake: Dividing the speed instead of the total distance by 2.

## Pause and Ponder

### Question 13

*3 marks · Numerical*

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 $4\text{ s}$ to return. What is the depth of the ocean at that location if the speed of sound in seawater is $1500\text{ m s}^{-1}$?

**Solution**

1. Given: Total time $t = 4\text{ s}$, Speed of sound $v = 1500\text{ m s}^{-1}$
2. Formula: $\text{Depth } d = \frac{v \times t}{2}$
3. Substitution: $d = \frac{1500\text{ m s}^{-1} \times 4\text{ s}}{2}$
4. Result: $3000\text{ m}$

**Answer:** The depth of the ocean is $3000\text{ m}$ (or $3\text{ km}$).

> Common mistake: Forgetting to divide the total distance travelled by 2.

## Revise, Reflect, Refine

### Question 1

*1 mark · MCQ*

Which observation best supports the idea that sound is a mechanical wave?

- Sound shows reflection
- Sound needs a medium to propagate
- Sound has frequency
- Sound carries energy

**Solution**

1. Waves that require a material medium for propagation are called mechanical waves.
2. Since sound cannot travel through a vacuum and needs a material medium, the observation that sound needs a medium to propagate best supports the idea that sound is a mechanical wave.
3. Final Answer: (ii) Sound needs a medium to propagate

**Answer:** (ii) Sound needs a medium to propagate

> Common mistake: Confusing properties like reflection or energy with the fundamental requirement of a medium for mechanical waves.

### Question 2

*1 mark · MCQ*

For a sound wave propagating in a medium, increasing its frequency will increase its

- wavelength
- speed
- number of compressions per second
- time period

**Solution**

1. Frequency is defined as the number of density oscillations (or compressions and rarefactions) at a fixed point per unit time.
2. Therefore, increasing the frequency of a sound wave will increase the number of compressions passing per second.
3. Final Answer: (iii) number of compressions per second

**Answer:** (iii) number of compressions per second

> Common mistake: Assuming frequency affects speed in all media, whereas speed depends only on the medium.

### Question 3

*1 mark · MCQ*

If 20 compressions pass a point in 4 seconds, the frequency is

- 80 Hz
- 5 Hz
- 10 Hz
- 0.2 Hz

**Solution**

1. Frequency is the number of oscillations per second, given by $\nu = \frac{\text{number of compressions}}{\text{time taken}}$.
2. Substituting the values, $\nu = \frac{20}{4\text{ s}} = 5\text{ Hz}$.
3. Final Answer: (ii) 5 Hz

**Answer:** (ii) 5 Hz

> Common mistake: Multiplying the values instead of dividing.

### Question 4

*3 marks · Short answer*

In a room, the reflected sound reaches the ear $0.05\text{ s}$ after its production. Will it produce an echo or reverberation? Justify your answer.

**Solution**

1. Reverberation occurs when sound reflections from surfaces arrive with a time difference less than $0.05\text{ s}$, while an echo requires a time interval of at least $0.1\text{ s}$.
2. Here, the reflected sound reaches the ear after $0.05\text{ s}$, which is less than the $0.1\text{ s}$ required to hear a distinct echo.
3. Thus, it will produce reverberation because the multiple reflections arrive too quickly to be distinguished separately from the original sound.

**Answer:** It will produce reverberation because the time interval of $0.05\text{ s}$ is less than the $0.1\text{ s}$ required for an echo.

> Common mistake: Confusing the time limits for echo ($0.1\text{ s}$) and reverberation ($0.05\text{ s}$).

### Question 5

*3 marks · Short answer*

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?

**Part (i)**

1. Wavelength is the distance between two consecutive crests.
2. From the given figure, wave (a) shows a larger distance between consecutive crests compared to wave (b).
3. Therefore, wave (a) has a greater wavelength.

Answer (i): Wave (a)

**Part (ii)**

1. Amplitude represents the maximum change in density from the average density, seen as the height of the crest from the average line.
2. Wave (b) has a smaller height of crests from the average line compared to wave (a).
3. Therefore, wave (b) has a smaller amplitude.

Answer (ii): Wave (b)

**Answer:** Wave (a) has greater wavelength and wave (b) has smaller amplitude.

> Common mistake: Mixing up wavelength (distance along X-axis) with amplitude (height along Y-axis).

### Question 6

*3 marks · Short answer*

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.

**Solution**

1. Frequency is directly proportional to the number of waves (oscillations) shown in a given distance, meaning higher frequency corresponds to shorter wavelength and more cycles.
2. Curve with the most oscillations in the given distance represents source A (maximum frequency), and the curve with the fewest oscillations represents source C (minimum frequency).
3. The intermediate curve represents source B.

**Answer:** Source A corresponds to the curve with the highest number of oscillations, source C to the fewest, and source B to the intermediate one.

> Common mistake: Relating high frequency with longer wavelength instead of shorter wavelength.

### Question 7

*3 marks · Short answer*

Draw a graph to represent a sound wave for which the density amplitude is 3 units and wavelength is $4\text{ cm}$.

**Solution**

1. Draw a Cartesian coordinate system with the x-axis representing distance in cm and the y-axis representing density.
2. Mark the average density as a horizontal dashed line on the y-axis.
3. Draw a sinusoidal wave where the peak (crest) is 3 units above the average density line and the trough is 3 units below it.
4. Mark the horizontal distance between two consecutive crests or troughs as 4 cm to represent the wavelength.

**Answer:** A sinusoidal graph with a peak amplitude of 3 units and a wavelength of 4 cm.

> Common mistake: Confusing density amplitude with displacement amplitude or mislabeling the axes.

### Question 8

*3 marks · Short answer*

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**

1. Sound is a mechanical wave that requires a material medium to propagate.
2. Outer space is a vacuum, meaning it lacks any medium for sound waves to travel.
3. Therefore, the depiction of sound in space is scientifically incorrect, while light can travel through a vacuum.

**Answer:** The error is that sound cannot travel in the vacuum of space because it is a mechanical wave requiring a medium, whereas light can travel through a vacuum.

> Common mistake: Forgetting that sound is a mechanical wave.

### Question 9

*3 marks · Numerical*

A source produces a sound wave of wavelength $3.44\text{ m}$. If the wave travels with a speed of $344\text{ m s}^{-1}$ find its time period.

**Solution**

1. Given: Wavelength $\lambda = 3.44\text{ m}$, Speed of wave $v = 344\text{ m s}^{-1}$
2. Formula: Speed $v = \nu \lambda$, and frequency $\nu = \frac{1}{T}$
3. Substitution: Frequency $\nu = \frac{v}{\lambda} = \frac{344\text{ m s}^{-1}}{3.44\text{ m}} = 100\text{ Hz}$
4. Result: Time period $T = \frac{1}{\nu} = \frac{1}{100\text{ Hz}} = 0.01\text{ s}$

**Answer:** $0.01\text{ s}$

> Common mistake: Forgetting that time period is the reciprocal of frequency.

### Question 10

*3 marks · Numerical*

A ship searching for a sunken ship sent a sonar signal and detected an echo after $5\text{ s}$. If ultrasonic wave travels at $1525\text{ m s}^{-1}$ in seawater, approximately how far down in the ocean is the wreckage of the sunken ship located?

**Solution**

1. Given: Time taken for echo $t = 5\text{ s}$, Speed of ultrasonic wave $v = 1525\text{ m s}^{-1}$
2. Formula: Distance $d = \frac{v \times t}{2}$
3. Substitution: $d = \frac{1525\text{ m s}^{-1} \times 5\text{ s}}{2}$
4. Result: $d = 3812.5\text{ m}$

**Answer:** $3812.5\text{ m}$

> Common mistake: Dividing the speed instead of the total distance travelled to and fro by 2.

### Question 11

*3 marks · Numerical*

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 $40\text{ kHz}$) which is reflected by the obstacle. When the warning beep starts sounding at a distance of $1.2\text{ m}$ 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 $345\text{ m s}^{-1}$.

**Solution**

1. Given: Distance to obstacle $d = 1.2\text{ m}$, Speed of ultrasonic wave $v = 345\text{ m s}^{-1}$
2. Formula: $\text{Total distance travelled} = 2 \times d = 2 \times 1.2\text{ m} = 2.4\text{ m}$
3. Formula: $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$
4. Substitution: $t = \frac{2.4\text{ m}}{345\text{ m s}^{-1}}$
5. Result: $t \approx 0.00696\text{ s}$ or $6.96\text{ ms}$

**Answer:** $6.96\text{ ms}$

> Common mistake: Dividing only the one-way distance by speed instead of the total distance to the obstacle and back.

### Question 12

*3 marks · Numerical*

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

**Solution**

1. Given: Distance $d = 1720\text{ m}$, Speed at $22\text{ °C}$ $v_1 = 344\text{ m s}^{-1}$, Speed at $0\text{ °C}$ $v_2 = 331\text{ m s}^{-1}$
2. Formula: $t_1 = \frac{d}{v_1}$ and $t_2 = \frac{d}{v_2}$
3. Substitution: $t_1 = \frac{1720\text{ m}}{344\text{ m s}^{-1}} = 5\text{ s}$
4. Substitution: $t_2 = \frac{1720\text{ m}}{331\text{ m s}^{-1}} \approx 5.196\text{ s}$
5. Result: Extra time = $t_2 - t_1 = 5.196\text{ s} - 5\text{ s} = 0.196\text{ s}$

**Answer:** $0.196\text{ s}$

> Common mistake: Subtracting speeds directly instead of calculating the times taken at both temperatures.

### Question 13

*3 marks · Numerical*

The variation of density of medium for a sound wave propagating with a speed of $340\text{ m s}^{-1}$ is shown in Fig. 10.32. Calculate the wavelength and frequency of the sound wave.

**Solution**

1. Given: Speed of sound $v = 340\text{ m s}^{-1}$, total distance for the given waves in Fig. 10.32 is $8\text{ cm} = 0.08\text{ m}$.
2. From the figure, the given length contains 2 complete waves, so wavelength $\lambda = \frac{0.08\text{ m}}{2} = 0.04\text{ m}$.
3. Formula: $v = \nu \times \lambda$
4. Substitution: $340\text{ m s}^{-1} = \nu \times 0.04\text{ m}$
5. Result: $\nu = \frac{340}{0.04} = 8500\text{ Hz} = 8.5\text{ kHz}$

**Answer:** Wavelength $\lambda = 0.04\text{ m}$, frequency $\nu = 8500\text{ Hz}$

> Common mistake: Confusing the total distance shown in the graph with a single wavelength.

### Question 14

*3 marks · Numerical*

The graphical representation of two sound waves A and B propagating at the same speed of $345\text{ m s}^{-1}$ is shown in Fig. 10.33. What is the wavelength of each of them? Also, calculate their frequencies.

**Part (i)**

1. From Fig. 10.33, wave A completes one full wave (one wavelength $\lambda$) in a distance of $5.0\text{ cm}$ (or $0.05\text{ m}$).
2. Wave B completes two full waves in $5.0\text{ cm}$, so its wavelength is $\lambda_B = \frac{5.0\text{ cm}}{2} = 2.5\text{ cm}$ (or $0.025\text{ m}$).
3. Therefore, wavelength of wave A is $5\text{ cm}$ ($0.05\text{ m}$) and wavelength of wave B is $2.5\text{ cm}$ ($0.025\text{ m}$).

Answer (i): Wavelength of wave A = $0.05\text{ m}$; Wavelength of wave B = $0.025\text{ m}$

**Part (ii)**

1. Using the relation between speed, wavelength, and frequency: $v = \lambda \nu$, we get frequency $\nu = \frac{v}{\lambda}$.
2. Given speed $v = 345\text{ m s}^{-1}$. For wave A with $\lambda_A = 0.05\text{ m}$, frequency $\nu_A = \frac{345\text{ m s}^{-1}}{0.05\text{ m}} = 6900\text{ Hz}$.
3. For wave B with $\lambda_B = 0.025\text{ m}$, frequency $\nu_B = \frac{345\text{ m s}^{-1}}{0.025\text{ m}} = 13800\text{ Hz}$.

Answer (ii): Frequency of wave A = $6900\text{ Hz}$; Frequency of wave B = $13800\text{ Hz}$

**Answer:** Wave A: wavelength = 5 cm, frequency = 6900 Hz; Wave B: wavelength = 2.5 cm, frequency = 13800 Hz

> Common mistake: Confusing the distance for one complete wave with the total distance shown on the graph.

### Question 15

*3 marks · Numerical*

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?

**Solution**

1. Given: Sound sources at A (air) and B (water) travel to the cliff and back over the same distance $d$.
2. Let the time taken by sound in water to return be $t_B = t$. Then the time taken in air is $t_A = 4.5 t$.
3. Distance to the cliff $d = \frac{v_{\text{air}} \times t_A}{2} = \frac{v_{\text{water}} \times t_B}{2}$.
4. Substitution: $v_{\text{air}} \times (4.5 t) = v_{\text{water}} \times t$
5. Result: $\frac{v_{\text{air}}}{v_{\text{water}}} = \frac{1}{4.5} = \frac{2}{9}$

**Answer:** Ratio of the speed of sound in air to that in water is $2:9$

> Common mistake: Inverting the ratio of speeds by confusing air and water times.

## Frequently asked questions

### How many total questions and exercises are included in the new NCERT Class 9 Science Chapter 10 on Sound Waves?

This chapter for the 2026-27 session based on the new NCF 2023 book includes 2 'Think It Over' questions, 10 'Pause and Ponder' questions, and 15 'Revise, Reflect, Refine' questions. SwaVid's free PDF and step-by-step solutions for all these questions are available on this page only.

### Which important topics and concepts do the practice questions cover in this chapter?

The questions comprehensively cover vital concepts such as the propagation and mechanical nature of sound waves, the necessity of a medium, density variations in graphical representations, and characteristics like frequency, time period, and wavelength. Other key topics include echo, reverberation, sonar depth calculation, and echolocation using ultrasonic waves.

### What are the hardest question types in Class 9 Science Chapter 10 and how should students approach them?

Numerical problems involving speed of sound in different media, time differences through steel and air, minimum distance for echoes, and sonar depth calculations are often found challenging. Students should approach them by carefully noting down the given values, applying relevant formulas step by step, and ensuring proper unit conversions.

### How can students write answers to numerical and conceptual questions to score full marks in exams?

To secure full marks in conceptual questions about sound wave characteristics or graphical representations, students must use clear scientific terminology and mention underlying principles like particle motion and energy transfer. For numerical problems, explicitly stating the given parameters, showing the formula substitution, and writing final answers with correct units is essential.

### Is a free PDF of NCERT solutions for Class 9 Science Chapter 10 available for download?

Yes, complete step-by-step solutions and the free PDF for this chapter adhering to the new NCERT book for the 2026-27 session are provided on this page only by SwaVid. Students can easily access these resources to practice different question types including MCQs, assertion-reason, and numerical problems.

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- [Class 9 Science chapters](https://www.swavid.com/science/class/9)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student