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Class 9 Science Chapter 10, Sound Waves: NCERT question answers for every Think It Over, Activity, Pause & Ponder, Worked Example, What If, Revise Reflect Refine and Journey Beyond question (CBSE 2026-27), with full working for every numerical. Keep your Class 9 Science chapter 10 notes handy and check each answer here.
Sound Waves asks how vibrations become something we can hear, and these Class 9 Science Chapter 10 NCERT solutions cover how sound travels as compressions and rarefactions, the wave equation v = λν numericals, the difference between loudness and pitch, echo distance and SONAR depth problems, and the uses of ultrasonic and infrasonic waves, with every numerical worked step by step.
Last updated 20 October 2026 · Reviewed by Suchita Arora, 13+ years teaching CBSE Maths & Science
Sound Waves: Characteristics and Applications takes sound from something you simply hear to something you can measure. It builds up how sound is produced by vibrations, why it needs a medium to travel as a longitudinal mechanical wave of compressions and rarefactions, and how that wave is described using wavelength, frequency, time period, amplitude and speed, connected by v = λν. It closes with echo and reverberation, and how ultrasonic and infrasonic waves power SONAR, echolocation and medical imaging. Every question is solved here, section by section, as complete Class 9 Science Chapter 10 NCERT solutions with answers.
Vibration as the source of every sound, why a medium (solid, liquid or gas) is essential, and how compressions and rarefactions form a longitudinal wave.
Wavelength, frequency, time period, amplitude and speed — and the one equation, v = λν, that ties every numerical in this chapter together.
Echo, reverberation, pitch and loudness, plus ultrasonic and infrasonic waves behind SONAR, echolocation and medical ultrasound.
Sound is a mechanical wave and needs a material medium (solid, liquid or gas) to travel.
Outer space is a near vacuum, so there is no medium between the astronauts.
So the sound of their voices or of the metal clanking cannot travel from one astronaut to the other.
They talk to each other through special communication devices (radio) fitted in their spacesuits. Radio waves do not need a medium.
A bat sends out short bursts of ultrasonic waves (frequency above 20 kHz).
These waves strike the prey or obstacles and get reflected back as echoes.
The bat senses the echoes. The time taken by the echo and its direction tell the bat the position of the prey or obstacle.
This method of locating objects using reflected sound is called echolocation.
Step 3: Yes, a sound is heard.
Step 4: Yes, the rubber band is seen vibrating (moving to and fro rapidly).
Step 5: No. When the vibration stops, the sound also stops.
Step 6: Yes, the sound changes. A more stretched (tighter) band vibrates faster and gives a higher pitch (shriller sound). A loosened band vibrates slower and gives a lower pitch.
Step 7: Yes, sound is still produced, but it is fainter, not as loud as before. The box is not there to strengthen the sound.
Step 3: Yes, a clear sound is heard from the vibrating tuning fork.
Step 4: Yes, the vibrating prong makes waves and splashes in the water.
Step 5: Yes, the sound is heard in every orientation. This shows that sound spreads in all directions from the source.
Step 1: The sound is heard, but it is faint.
Step 2: Yes, the sound is heard and it is louder and clearer through the desk.
Step 2: A clear ringing sound is heard in air.
Step 3: Yes, the sound is heard again, though it may be a little different.
The sound has reached our ear after travelling through water and then air.
Step 4: Yes. A region where the turns are closer together moves along the slinky towards the friend.
Step 5: Yes. A series of regions with closely spaced turns and regions with spread out turns is produced.
These regions move along the whole length of the slinky.
The mark does not travel along with the disturbance. It only moves back and forth about its own position, parallel to the direction in which the disturbance travels.
Step 4: Yes. The grains move and jump on the sheet.
Step 5: A louder sound makes the grains jump higher, and a softer sound makes them move less. The effect is seen with all kinds of grains.
Reason: The sound travels through the air and reaches the sheet. It makes the sheet vibrate, and the vibration moves the grains.
Sample readings (the exact values depend on the pitch chosen for Sa):
| Note | Frequency (Hz) | Ratio to Sa |
|---|---|---|
| Sa | 240 | 1 |
| Re | 270 | 1.125 |
| Ga | 300 | 1.25 |
| Ma | 320 | 1.33 |
| Pa | 360 | 1.5 |
| Dha | 400 | 1.67 |
| Ni | 450 | 1.875 |
| Sa (upper) | 480 | 2 |
Step 2: The frequency is lowest for the first Sa and keeps increasing for Re, Ga, Ma, Pa, Dha and Ni.
Step 4: The ratios are fixed numbers, which do not depend on the pitch of Sa. The upper Sa has a frequency exactly double that of the lower Sa (ratio 2).
Step 5: The values for the voice are close to those of the app, but slightly different because a voice cannot hold an exact frequency.
Step 2: A low, deep sound is heard at 100 Hz.
Step 3: As the frequency increases, the sound becomes higher in pitch, from deep to sharp and shrill. The pitch rises with frequency.
Step 4: At 50 Hz the sound is very deep. Near 20 Hz it is hardly heard, and below about 20 Hz nothing can be heard.
Sound is produced whenever an object vibrates. Some ways are:
| Way of producing sound | Example |
|---|---|
| Plucking a stretched string | Sitar, guitar, rubber band |
| Striking an object | Tabla, drum, bell, tuning fork |
| Blowing air in a hollow pipe | Bansuri (flute) |
| Vibrating vocal cords | Talking, singing |
| Rubbing body parts | Crickets and grasshoppers rub wings or legs |
| Clapping hands | Hands striking each other |
Musical instruments are grouped by the part that vibrates:
| Instrument | Vibrating part |
|---|---|
| Sitar, veena, guitar, tanpura | Stretched strings |
| Tabla, mridangam, drum | Stretched membrane (drum head) |
| Bansuri (flute) | Air column inside the pipe |
| Harmonium | Metal reeds |
| Bell, cymbals | Metal body |
| Xylophone | Wooden or metal bars |
Assertion (A): True. As the air is pumped out, the sound becomes fainter, and in a near vacuum almost no sound is heard.
Reason (R): True. Sound is a mechanical wave and needs a medium to travel.
Since there is no air (medium) left, the sound cannot reach us. So R explains A.
Assertion (A): True. The compressions and rarefactions travel through the medium.
Reason (R): False. The particles do not move forward with the wave. They only vibrate to and fro about their mean positions.
The particles of air only vibrate about their mean positions. They do not travel from the fork to the ear, so (i) and (iv) are wrong.
The material of the tuning fork stays with the fork, so (iii) is wrong.
What reaches the ear is the disturbance (compressions and rarefactions), which carries energy.
A region where the dots are crowded has a high density. It is a compression (C), and it matches a crest of the graph.
A region where the dots are spread out has a low density. It is a rarefaction (R), and it matches a trough of the graph.
On the graph, distance is on the x-axis and density is on the y-axis. The dashed line is the average density.
Wave (a): the dense and sparse regions are close together (shorter wavelength), so the curve (c) has more waves.
Wave (b): the dense and sparse regions are farther apart (longer wavelength), so the curve (d) has fewer waves.
Observation: Yes, the thin rubber band vibrates faster than the thick rubber band.
The thin band makes more vibrations in one second. So its frequency is higher, and it gives a higher pitch (shriller sound).
Time period is the time for one vibration, T = 1/ν. A higher frequency means a shorter time period.
| Rubber band | Speed of vibration | Frequency | Time period |
|---|---|---|---|
| Thin | Faster | Higher | Shorter |
| Thick | Slower | Lower | Longer |
Given: frequency ν = 20 Hz
To find: number of oscillations in 1 minute
Frequency 20 Hz means 20 oscillations in every second.
1 minute = 60 s
Wavelength is the distance between two consecutive crests.
From the graph, the crests are at 0.75 cm and 3.75 cm.
Also, one full wave starts at 0 and ends at 3.0 cm, so λ = 3.0 cm.
Half of the wavelength:
| State | Substance/Medium | Approximate speed |
|---|---|---|
| Solid | Steel | 5000 m s⁻¹ |
| Liquid | Water | 1500 m s⁻¹ |
| Gas | Air | 340 m s⁻¹ |
Given: steel = 5000 m/s, water = 1500 m/s, air = 340 m/s
Given: distance = 340 m, speed in steel = 5000 m/s, speed in air = 340 m/s
To find: time difference between the two sounds
Time = distance / speed
Through air:
Through steel:
Time difference:
0.932 s is more than 0.1 s.
Given: time t = 0.2 s, speed v = 343 m/s
To find: minimum distance of the reflecting surface
The sound goes to the surface and comes back, so the total distance covered = 2d.
Given: total time t = 4 s, speed v = 1500 m/s
To find: depth of the ocean
The signal goes down and comes back, so the time to reach the bottom = 4/2 = 2 s.
Given: number of oscillations = 10, time = 2 s
Frequency = number of oscillations / time taken
Time for one oscillation:
Given: speed v = 344 m/s; frequencies 20 Hz and 20 kHz = 20000 Hz
Using v = λ × ν, the wavelength is λ = v / ν.
Given: time delay t = 5 s, speed of sound v = 340 m/s
To find: distance to the lightning strike
Light reaches us almost at once, so the delay is the time taken by sound.
Given: speed v = 5000 m/s
Wavelength: from the graph, the distance between two consecutive crests is 50 m.
Frequency: using v = λ × ν
Time period:
Given: time t = 0.5 s, speed v = 340 m/s
To find: distance from the wall
The sound goes to the wall and comes back, so the total distance = 2d.
Given: total time = 0.90 s, speed v = 1530 m/s
To find: distance of the object
Time taken to reach the object = half of the total time:
Distance:
A musical note is a mixture of a fundamental frequency and higher frequencies (overtones). All of them leave the source together.
If the speed depended on frequency, the different frequencies would travel at different speeds and reach the listener at different times.
The sound would be blurred, especially for listeners sitting far from the stage. The notes of the singer and the instruments would also not stay in step.
Advantages:
We could detect ultrasonic sounds from machines and find faults early. We could sense the high-pitched calls of animals, and use sound to find objects in the dark, like bats do.
Disadvantages:
We would hear a lot of extra sound from machines, electronic devices and animals all the time. This would cause noise, disturb sleep and concentration, and could harm hearing.
Energy: The mechanical (kinetic) energy of a vibrating object is converted into sound energy. For example, the muscular energy of a person plucking a string or striking a drum.
Production: Sound is produced by vibrating objects, such as strings, membranes, air columns and vocal cords.
Reaching the ear: The vibrating source makes the particles of the medium (air) vibrate, which produces compressions and rarefactions. This disturbance travels through the medium and reaches the ear, where it makes the eardrum vibrate.
A vacuum has no matter (no medium), so there are no particles to carry the vibrations.
In the bell jar experiment, the sound becomes fainter as air is pumped out, and almost no sound is heard in a near vacuum, though the bell is seen ringing.
The spoons are inside the water, so the sound first travels through the water and then through the air to reach our ears.
We do hear the sound when the spoons are submerged. If liquids could not carry sound, no sound would have come out of the water.
Sound waves are mechanical waves. They travel by the vibration of the particles of a medium and their collisions with the neighbouring particles.
Without particles, there is nothing to vibrate and pass on the disturbance.
A mechanical wave is a wave that needs a material medium to propagate.
Reflection, frequency and carrying energy are properties of all waves, including light, which is not a mechanical wave.
Only the need for a medium shows that sound is a mechanical wave.
Speed depends only on the medium, so it stays the same. (ii) is wrong.
Since v = λν, with a fixed speed, a larger frequency gives a smaller wavelength. (i) is wrong.
Time period T = 1/ν decreases. (iv) is wrong.
Frequency is the number of oscillations (compressions) passing a point per second, so it increases.
Given: 20 compressions (20 oscillations) in 4 s
Each compression passing means one oscillation.
An echo is heard only when the reflected sound arrives at least 0.1 s after the original sound.
Here the time gap = 0.05 s, which is less than 0.1 s, so the ear cannot hear the reflected sound separately.
Instead, the multiple reflections from the walls of the room blend with the original sound and make it persist. This is reverberation.
Wavelength is the distance between two consecutive crests. Amplitude is the maximum height of the curve above the average density line.
In the same length of the x-axis, wave (a) has 3 waves and wave (b) has 6 waves.
The curve of (a) rises and falls less from the average line than that of (b).
Fewer waves in the same distance means longer waves.
Wave (a) has a smaller height above the average line.
Frequency is the number of waves per unit distance (or time). More waves in the same length means a higher frequency (a shorter wavelength).
The green curve has the most waves, the red curve has fewer, and the blue curve has the fewest.
| Curve | Number of waves | Frequency | Source |
|---|---|---|---|
| Green | Most | Maximum | A |
| Red | In between | In between | B |
| Blue | Fewest | Minimum | C |
Amplitude = 3 units, so the density rises 3 units above the average and falls 3 units below it.
Wavelength = 4 cm, so the crests are 4 cm apart.
Take distance (cm) on the x-axis and the change in density from the average on the y-axis.
Crests are at 1 cm and 5 cm (height +3), troughs are at 3 cm and 7 cm (height −3), and the curve crosses the average line at 0, 2, 4, 6 and 8 cm.
Error 1: Space is a near vacuum. Sound cannot travel without a medium, so no explosion sound can be heard.
Error 2: Light travels much faster than sound (300000 km/s compared with about 340 m/s in air). Even if there were air, the flash would be seen first and the sound would be heard later, not at the same time.
Given: wavelength λ = 3.44 m, speed v = 344 m/s
To find: time period T
In one time period, the wave travels one wavelength, so v = λ / T.
Given: total time = 5 s, speed v = 1525 m/s
To find: depth of the wreck
The time to reach the wreck = 5/2 = 2.5 s.
Given: distance of obstacle = 1.2 m, speed v = 345 m/s
To find: time taken to go to the obstacle and come back
The wave travels to the obstacle and back, so the total distance = 2 × 1.2 = 2.4 m.
Given: distance = 1720 m; speed at 22 °C = 344 m/s; speed at 0 °C = 331 m/s
To find: extra time taken at 0 °C
At 22 °C:
At 0 °C:
Extra time:
Given: speed v = 340 m/s
In Fig. 10.32, the arrow of length 8 cm runs from the centre of the first compression to the centre of the third compression.
From one compression to the next is one wavelength, so 8 cm covers 2 wavelengths.
Using v = λ × ν
Given: speed of both waves v = 345 m/s
From the graph, wave A completes 3 full waves in 7.5 cm, and wave B completes 1.5 waves in 7.5 cm. So:
Using ν = v / λ:
Let the distance from each source to the cliff be d. Each sound travels 2d (to the cliff and back).
Let v₁ = speed in air (source A) and v₂ = speed in water (source B).
Given: t_A = 4.5 × t_B
Cancel 2d from both sides:
Rearranging:
A sample article (points to include):
1. Effect of earphones: Studies by health bodies, such as the World Health Organization (WHO), report that listening to loud music for a long time through earphones can damage the ear and cause permanent hearing loss, especially in young people. Safe listening habits are to keep the volume below about 60 % of the maximum and to take breaks.
2. How hearing is tested: A hearing test called audiometry is done by an audiologist. The person listens to pure tones of different frequencies and loudness through headphones and signals when a tone is heard. The result is plotted as an audiogram.
3. Decibel ranges (hearing level, one commonly used scale):
| Degree of hearing loss | Hearing level (dB) |
|---|---|
| Mild | 26 to 40 |
| Moderate | 41 to 55 |
| Moderately severe | 56 to 70 |
| Severe | 71 to 90 |
| Profound | above 90 |
4. Government schemes: The scheme ADIP (Assistance to Disabled Persons for Purchase/Fitting of Aids and Appliances) of the Government of India supports hearing aids and other aids, and also covers cochlear implants for young children with severe hearing loss. The National Programme for Prevention and Control of Deafness (NPPCD) also works on screening and treatment. The eligibility rules change from time to time, so check the official websites before writing the final article.
Procedure: Make a cone from poster paper, with a small opening at the narrow end. Place the phone playing music at the narrow end and listen from the wide end. Then listen without the cone from the same distance.
Expected observation: With the cone, the sound is louder in front of the wide opening than without the cone.
Reason: The walls of the cone reflect the sound waves again and again, and send them forward in one direction, instead of letting them spread out in all directions. This is the principle of a megaphone.
Different shapes: A cone with a wider mouth or a longer cone directs the sound better. A shape that does not guide the sound forward, such as a flat sheet, makes little difference.
A curved surface behind the stage reflects the sound coming from the stage towards the audience.
A flat surface reflects sound in only one fixed direction, and the reflected sound may not reach all parts of the hall.
A curved ceiling and wall spread the reflected sound evenly over the whole audience, so even the people sitting at the back hear clearly.
The design also reduces unwanted echoes and reverberation, so the speech and music are clear.
Sample readings: distance = 250 m. The times noted: 0.72 s, 0.74 s, 0.73 s, 0.73 s and 0.73 s.
Average time:
(v) Speed of sound:
This is close to the typical value of 346 m/s. A small difference arises from errors in the reaction time of starting and stopping the stopwatch and in measuring the distance.
(vi) Why measure the time between seeing and hearing: Light travels at 300000 km/s, so we see the burst almost at the same instant it happens. The time between seeing and hearing is the time taken by sound to travel the distance. If we started the stopwatch when the friend gave the signal, the time taken by the signal would be added and the result would be wrong.
Effect of temperature: The speed of sound in air increases when the temperature increases. For example, it is about 331 m/s at 0 °C and about 344 m/s at 22 °C.
Effect of humidity: The speed of sound in air increases when the humidity increases.
Reason: At a higher temperature, the particles of air move faster, so the disturbance is passed on from particle to particle more quickly.
Sound is a vibration that becomes a travelling disturbance of compressions and rarefactions in a medium — carrying energy but not matter — and every property we can measure about it (wavelength, frequency, time period, amplitude, and speed, tied together by v = λν) explains what we hear as pitch and loudness, why sound reflects as echo or reverberation, and how frequencies beyond our own hearing, from a bat's ultrasonic call to a ship's SONAR ping, let us sense what our ears alone cannot.
Seven extra questions in the style of the textbook's own activities and examples, for independent practice once you've gone through everything above. Attempt each one on paper first, then tap to check.
Given: ν = 250 Hz, v = 340 m/s
To find: wavelength λ
Given: total time = 6 s, v = 1500 m/s
To find: depth of the sea
Time to reach the bottom = 6/2 = 3 s.
Loudness depends mainly on the amplitude of the wave. Both waves have the same amplitude, so they carry about the same energy, and they will sound about equally loud.
Pitch depends on frequency. The 2000 Hz sound has a higher frequency, so it has a higher pitch (sounds shriller), and the 200 Hz sound has a lower pitch (sounds deeper).
Given: T = 0.02 s
To find: frequency, and whether it is audible
The audible range is 20 Hz to 20000 Hz, and 50 Hz lies in this range.
Given: distances = 340 m and 510 m; v = 340 m/s
To find: time interval between the two echoes
Each sound goes to a cliff and back, so the total distance = 2d.
Echo from the first cliff:
Echo from the second cliff:
Time interval:
1 s is more than 0.1 s, so the two echoes are heard separately.
Given: ν = 512 Hz; speed in air = 340 m/s; speed in water = 1500 m/s
Using λ = v / ν:
The frequency is decided by the source (the tuning fork), so it stays the same at 512 Hz. The speed changes with the medium, so the wavelength also changes.
Large empty hall: It has hard, smooth walls, floor and ceiling that reflect sound well. Sound is reflected many times from these surfaces. Because the hall is large, the reflections reach us at slightly different times, so the sound persists after the clap stops (reverberation).
Small room with curtains and carpets: Soft, porous surfaces such as curtains and carpets absorb sound instead of reflecting it, so very little sound is reflected back. Also, in a small room the reflections come back so quickly that they cannot be separated from the original sound.
This is why concert halls use absorbing panels, curtains and upholstered chairs to control reverberation.
Browse the full list of Class 9 Science (Exploration) NCERT Solutions chapters, visit the Class 9 Science hub, or revise with the Class 9 Science Notes, or test yourself with the free Class 9 Science Sample Paper and Marking Scheme.
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