NCERT Solutions for Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications 2026-27

Revise, Reflect, Refine (NCERT Textbook Page No. 204)

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

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

Sound is called a mechanical wave because it cannot travel without a material medium. It needs air, water, solids, or any other medium to move from one place to another.

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

Answer:
(iii) number of compressions per second.

Frequency means the number of vibrations produced in one second. So, when frequency increases, the number of compressions and rarefactions produced per second also increases. In the same medium, the speed of sound remains almost constant, while wavelength and time period decrease.

Question 3. If 20 compressions pass a point in 4 seconds, the frequency is
(i) 80 Hz
(ii) 5 Hz
(iii) 10Hz
(iv) 0.2 Hz

Answer:
(ii) 5 Hz

Frequency is the number of compressions passing a point per second.

Frequency = Number of compressions / Time
Frequency = 20 / 4 = 5 Hz

So, the frequency is 5 Hz.

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.

Answer:
It will produce reverberation, not a clear echo.

For a distinct echo to be heard, the reflected sound must reach the ear at least 0.1 s after the original sound. Here, the reflected sound reaches after only 0.05 s, which is less than 0.1 s. Therefore, the original and reflected sounds overlap and are heard as 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?

Answer:
(i) Fig. 10.30 (a) has a greater wavelength because the distance between two consecutive crests or compressions is larger than in Fig. 10.30 (b).

(ii) Fig. 10.30 (a) has a smaller amplitude because the height of the wave from the mean position is less than that of Fig. 10.30 (b).

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.

Answer:
Frequency depends on the number of oscillations in a given distance or time.

• The curve with the maximum number of oscillations represents the highest frequency, so it should be marked A.

• The curve with a moderate number of oscillations represents the middle frequency, so it should be marked B.

• The curve with the least number of oscillations represents the minimum frequency, so it should be marked C.

Thus, label the most closely spaced wave as A, the middle one as B, and the least closely spaced wave as C.

Question 7. Draw a graph to represent a sound wave for which the density amplitude is 3 units, and wavelength is 4 cm.

Answer:
To draw the sound wave:

• Mark the mean density line as the central horizontal line.

• Take the maximum density variation as +3 units and minimum as -3 units from the mean line.

• Take the distance between two consecutive crests or compressions as 4 cm.

• Draw a smooth wave showing alternate compressions and rarefactions.

So, the graph should have an amplitude of 3 units and a wavelength of 4 cm.

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?

Answer:
The scene is scientifically incorrect for two reasons.

First, sound cannot travel in space because space is a vacuum and sound needs a material medium to propagate. So, the explosion should not be heard.

Second, light travels much faster than sound. Even in a medium where sound can travel, light and sound would not reach the observer at the same time.

Question 9. A source produces a sound wave of wavelength 3.44 m. If the wave travels with a speed of 344 ms-1 find its time period.

Answer:
Given:
Wavelength, λ = 3.44 m
Speed, v = 344 m/s

Using the formula:

v = λ / T

So,

T = λ / v
T = 3.44 / 344
T = 0.01 s

Therefore, the time period of the sound wave is 0.01 s.

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 ms-1’ in seawater, approximately how far down in the ocean is the wreckage of the sunken ship located?

Answer:
Given:
Speed of ultrasonic wave in seawater = 1525 m/s
Time taken for echo = 5 s

The sound travels from the ship to the wreckage and then returns. So, the total distance travelled is:

Distance = Speed × Time
Distance = 1525 × 5
Distance = 7625 m

This is the total to-and-fro distance. Therefore, the depth of the wreckage is:

Depth = 7625 / 2
Depth = 3812.5 m

So, the wreckage is approximately 3812.5 m deep, or about 3813 m below the surface.

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 ms-1.

Answer:
Given:
Distance from obstacle = 1.2 m
Speed of ultrasonic wave = 345 m/s

The wave travels to the obstacle and comes back. So, total distance travelled is:

Total distance = 2 × 1.2 = 2.4 m

Time = Distance / Speed
Time = 2.4 / 345
Time ≈ 0.00696 s

So, the time taken is approximately 0.007 s, or 7 milliseconds.

Question 12. The speed of sound in air is about 331 ms-1 at O °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 oc to 0 °C? Assume that all other conditions remain unchanged.

Answer:
Given:
Speed of sound at 22°C = 344 m/s
Speed of sound at 0°C = 331 m/s
Distance = 1720 m

Time taken at 22°C:

t₁ = 1720 / 344 = 5 s

Time taken at 0°C:

t₂ = 1720 / 331 ≈ 5.20 s

Extra time taken:

Extra time = 5.20 - 5
Extra time = 0.20 s

Therefore, the sound of thunder will take about 0.2 seconds extra at 0°C.

Question 13. The variation of density of medium for a sound wave propagating with a speed of 340 ms-1’ is shown in Fig. 10.32. Calculate the wavelength and frequency of the sound wave.

Answer: From the graph, the distance between two consecutive compressions or rarefactions gives the wavelength.

Given wavelength = 4 cm
= 0.04 m

Speed of sound, v = 340 m/s

Using the formula:

v = fλ

f = v / λ
f = 340 / 0.04
f = 8500 Hz

Therefore, the wavelength is 0.04 m, and the frequency is 8500 Hz.

Question 14. The graphical representation of two sound waves A and B propagating at the same speed of 345 ms-1 is shown in Fig. 10.33. What is the wavelength of each of them? Also, calculate their frequencies.

Answer: From the graph:

Wavelength of wave A = 2.5 cm = 0.025 m
Wavelength of wave B = 5 cm = 0.05 m

Speed of both waves = 345 m/s

Using the formula:

v = fλ

For wave A:

fA = 345 / 0.025
fA = 13,800 Hz

For wave B:

fB = 345 / 0.05
fB = 6,900 Hz

Therefore:

• Wavelength of wave A = 0.025 m, frequency = 13,800 Hz

• Wavelength of wave B = 0.05 m, frequency = 6,900 Hz

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?

Answer: Let the speed of sound in air be vA and the speed of sound in water be vB.

The distance travelled by both sound waves is the same. Time taken is inversely proportional to speed.

Given:

tA = 4.5 tB

So,

vA / vB = tB / tA
vA / vB = 1 / 4.5
vA / vB = 2 / 9

Therefore, the ratio of the speed of sound in air to the speed of sound in water is: 2 : 9

Important Formulas in Chapter 10 Sound Waves Class 9

• Speed of sound: v = λ × f

• Frequency: f = 1/T (number of oscillations per second)

• Time period: T = 1/f

• Distance in echo problems: Total distance = Speed × Time; Depth/Distance = (Speed × Time)/2

• Audible range of human hearing: 20 Hz to 20,000 Hz

Benefits of Vedantu’s NCERT Solutions for Class 9 Science Chapter 10

• All answers follow the latest CBSE 2026-27 syllabus and NCERT guidelines.

• Step-by-step solutions to numericals on speed, frequency, wavelength, and echo make problem-solving easy.

• Answers are written in simple language, ideal for quick understanding and last-minute revision.

• Helps students frame answers as per the CBSE marking scheme to score full marks.

• FREE PDF download lets you study offline, anytime and anywhere.

CBSE Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications Study Materials

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