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NCERT Solutions for Class 9 Science Chapter 10: Sound Waves – Characteristics and Applications
Class 9 Science Chapter 10: Sound Waves – Characteristics and Applications explains how sound is produced, how it travels as a mechanical wave, and how its properties power real-world technologies like SONAR and ultrasound imaging. Curated by ALLEN Experts and aligned with the new NCERT Exploration textbook (2026–27), these chapter-wise NCERT Solutions cover production and propagation of sound, wave characteristics (wavelength, frequency, time period, amplitude, and speed); pitch and loudness; reflection of sound, echo; reverberation, and echolocation — with step-by-step answers to every in-text and end-of-chapter question, plus important questions for focused practice.
1.0Download NCERT Solutions for Class 9 Science Chapter 10 – Sound Waves PDF
Download the NCERT Solutions for Class 9 Science Chapter 10 PDF to access comprehensive answers to all NCERT exercise questions in one place. Prepared by subject experts, these solutions include concept-based explanations, diagrams wherever required, and exam-oriented answers aligned with the latest CBSE curriculum.
2.0Learning Outcomes : NCERT Class 9 Science Chapter 10 Solutions
- Understand the Production of Sound: Explain how sound is produced by the vibration of objects and identify the source of sound in different situations.
- Understand the Propagation of Sound: Explain why sound needs a material medium to travel and cannot pass through a vacuum.
- Explain Sound as a Longitudinal Wave: Describe how sound travels through compressions and rarefactions in the direction of wave propagation.
- Define Key Wave Characteristics: Understand wavelength, frequency, time period, amplitude, intensity, and speed of a sound wave.
- Differentiate Pitch and Loudness: Relate pitch to frequency and loudness to amplitude of a sound wave.
- Compare Speed of Sound in Different Media: Explain how sound travels at different speeds through solids, liquids, and gases.
- Understand Reflection of Sound: Explain how sound waves reflect off surfaces, similar to light waves.
- Explain Echo and Reverberation: Understand the conditions required to hear a distinct echo and how reverberation affects sound in enclosed spaces.
- Understand the Range of Hearing: Differentiate audible, infrasonic, and ultrasonic sound ranges for humans and animals.
- Explore Applications of Ultrasound: Explain the use of ultrasonic waves in SONAR, echolocation, medical imaging, and industrial cleaning.
- Solve NCERT Exercise Problems: Confidently attempt and solve all in-text activities and end-of-chapter numerical and conceptual questions.
- Build a Foundation for Advanced Topics: Develop conceptual clarity that supports the study of waves and oscillations in higher classes.
3.0Detailed NCERT Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications Solutions
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, astronauts cannot talk to each other directly through space, nor can they hear sounds of metal clanking the way they do on Earth. Sound needs a material medium (such as air, water, or solids) to travel. Outer space is essentially a vacuum, so sound cannot propagate through it. Therefore:
- If one astronaut speaks, the other cannot hear the voice through space.
- The sound of metal tools striking the space station cannot travel through the vacuum to their ears. However, astronauts can communicate using radio transmitters built into their spacesuits. Radio waves are electromagnetic waves and can travel through a vacuum.
2. How do most bats use sound to locate their prey in the dark at night?
Solution
Most bats use a technique called echolocation.
- The bat emits high-frequency sounds (ultrasonic waves).
- These sound waves travel through the air and strike objects such as insects, trees, or walls.
- The waves are reflected back as echoes.
- By analyzing the echoes, the bat can determine the position, distance, size, shape, and movement of its prey. This allows bats to fly and hunt accurately even in complete darkness. Most bats locate their prey at night by emitting ultrasonic sounds and listening to the echoes reflected from nearby objects. This process is called echolocation.
3. Explore various ways of producing sound.
Solution
Sound is produced whenever an object is forced to move back and forth rapidly, creating a mechanical disturbance in the surrounding medium.
- Plucking and Bowing (Vibrating Strings) Mechanical force disturbs a tightly stretched string, causing it to vibrate and push the surrounding air molecules.
- Plucking: Pulling a string out of alignment and releasing it suddenly (e.g., Guitar, Sitar, Harp).
2. Striking and Beating (Vibrating Surfaces)
An impact transfers kinetic energy directly to a surface, causing it to flex and vibrate at its natural frequencies.
- Membranes: Hitting a stretched skin makes the surface ripple up and down, displacing a large volume of air (e.g., Drums, Tabla, Dholak).
- Solid Bodies: Striking rigid materials like metal, wood, or clay forces the entire structure into a complex vibration pattern (e.g., Xylophone, Cymbals, Ghatam).
- Blowing Air (Vibrating Air Columns) Directing a stream of air sets up standing wave patterns inside a hollow chamber or pipe.
- Lip Buzzing: The player tightens their lips and blows through them, causing the lips to act like a double reed to vibrate the air inside a metal tube (e.g., Trumpet, Trombone).
4. Make a list of different types of musical instruments and identify their vibrating parts which produce sound.
Solution
Here is the list of the different types of musical instruments and their specific vibrating parts:
- Stringed Instruments: - These instruments produce sound when their strings are plucked, bowed, or struck.
- Vibrating Part: Stretched strings
- Examples: Violin, guitar, sitar, piano, cello, harp.
- Wind Instruments: - These instruments produce sound when a column of air inside them is set into vibration.
- Vibrating Part: Air column (inside the hollow tube)
- Examples: Flute, trumpet, saxophone, clarinet, mouth organ, shehnai.
- Percussion Instruments: - These instruments produce sound when a stretched skin or membrane is struck with hands or sticks.
- Vibrating Part: Stretched membrane (skin)
- Examples: Tabla, dholak, drums, mridangam, bongo.
- Idiophones: - These instruments do not have strings or membranes; the entire body of the instrument vibrates when struck, shaken, or scraped.
- Vibrating Part: The entire body of the instrument
- Examples: Manjira (cymbals), ghatam (earthen pot), jal tarang (water-filled bowls), xylophone.
5. Assertion (A): We cannot hear 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: (i) Both A and R are true, but R is not the correct explanation of A. (ii) Both A and R are true, and R is the correct explanation of A. (iii) A is true, but R is false. (iv) A is false, but R is true.
Solution
(ii) Both A and R are true, and R is the correct explanation of A. Assertion (A) is true: In the classic bell jar experiment, pumping air out creates a partial vacuum. As the air density decreases inside the jar, the sound of the ringing bell becomes fainter until it can no longer be heard. Reason is the correct explanation of Assertion: Because sound strictly requires a material medium to propagate, removing the air (medium) prevents the sound vibrations from traveling from the bell to the outside walls of the jar, thereby making it inaudible.
5. 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: (i) Both A and R are true, but R is not the correct explanation of A . (ii) Both A and R are true, and R is the correct explanation of A. (iii) A is true, but R is false. (iv) A is false, but R is true
Solution
(iii) A is true, but R is false.
Assertion (A) is true: In a longitudinal wave (like a sound wave), the wave travels through a material medium by creating alternating regions of high pressure (compressions) and low pressure (rarefactions). These regions propagate or move forward continuously through the medium. Reason ( R ) is false: The individual particles of the medium do not travel forward with the wave. Instead, they only oscillate or vibrate back and forth about their fixed mean equilibrium positions. They transfer energy to neighbouring particles and return to their original spots, meaning only the disturbance (energy) moves forward, not the matter itself.
6. When sound travels from a tuning fork to your ear, which of the following actually reaches your ear? (i) Air particles near the tuning fork (ii) Energy carried by sound waves (iii) The tuning fork material (iv) A continuous stream of compressed air
Solution
(ii) Energy carried by sound waves
When a tuning fork vibrates, it displaces the air particles immediately surrounding it. These particles do not travel all the way to your ear. Instead, the particles simply vibrate back and forth about their fixed, mean positions. They bump into neighbouring particles, pass along the kinetic energy, and then return to their original starting point.
This continuous process creates regions of high pressure (compressions) and low pressure (rarefactions). It is this disturbance or the energy that propagates through the air and ultimately reaches your ear to produce the sensation of hearing.
7. The variation of density of the medium for two sound waves is shown in figure (a) and (b). Label compression and rarefaction by C and R on it. In the graph given in figure (c) and (d), label the axes and draw the curves corresponding to figure (a) and (b).
Drawing graph to represent the variation of density of the medium for two sound waves
Solution
To label the density variation figures and translate them into pressure/density graphs, follow these specific instructions for the regions and axes: Labelling Compression and Rarefaction on figure (a) and (b)
- Compression (C): Label C on the regions where the vertical lines (representing particles) are tightly packed together. These are regions of high density and pressure.
- Rarefaction (R): Label R on the regions where the vertical lines are spread far apart. These are regions of low density and pressure.
- Labelling the Axes for Graphs (c) and (d)
- Y-Axis (Vertical): Label this axis as Density (or Pressure).
- X-Axis (Horizontal): Label this axis as Distance (or Time).
- Centre Baseline: Draw a horizontal dashed line through the middle to represent the Mean Density (normal atmospheric pressure).
- Drawing the Curves
- Above the baseline (Crest): Draw the peak of the curve directly corresponding to the Compression (C) region from the figure above it.
- Below the baseline (Trough): Draw the valley of the curve directly corresponding to the Rarefaction ( R ) region from the figure above it.
- 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
Yes, a thin rubber band vibrates faster than a thick rubber band, which produces a sound with a higher pitch.
- Frequency: The frequency of the sound (the number of vibrations per second) produced by the thin rubber band is higher than that of the thick rubber band.
- Time Period: The time period (the time taken to complete one full vibration) of the thin rubber band's sound is shorter than that of the thick rubber band. This is because time period is inversely proportional to frequency. Time period = 1/Frequency
9. 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
we have the following data: frequency ( f ) time minute seconds frequency number of oscillations (n) = frequency (f) × time taken ( t )
10. For the sound wave represented by the graph shown in Fig. 10.19, what is half of its wavelength?
Graphical representation of a sound wave
Solution
A complete wavelength is defined as the distance between two consecutive compressions (crests) or two consecutive rarefactions (troughs).
One complete wave cycle spans from 0 to 3 cm
Therefore, the wavelength ( ) To find half of this wavelength, divide the total wavelength by 2 :
Half of wavelength
11. Table 10.1 shows the speed of sound in a few media at atmospheric pressure.
Speed of sound in different media
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.
Solution
The values provided in Table 10.1: Speed of sound in Air (Vair) Speed of sound in Water Speed of sound in Steel ( (i) Ratio of the speed of sound in water with respect to air (approx.) (ii) Ratio of the speed of sound in steel with respect to water (approx.) The speed of sound in steel is approximately 3.3 times faster than in water.
12. Two friends are standing along a steel fence at a distance of 340 m from each other figure. 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 s to be heard separately.)
Solution
From table 10.1: Speed of sound in air (Vair) Speed of sound in steel ( Distance between friends (d) = 340 m The formula to calculate time is: time Time taken through air ( ): Time taken through steel ( ): Time Difference
13. 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 .
Solution
we have the following data: Speed of sound ( v ) Minimum time delay ( t ) An echo involves the sound wave travelling to the reflecting surface and bouncing back to the source. The total distance covered during this time is twice the actual distance to the surface (2d).
14. 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 ?
Solution
we have the following data: Speed of sound in seawater ( v ) Total time taken for the round trip ( t ) Let the depth of the ocean Speed or 3 km
15. 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
Solution
(i) Sound needs a medium to propagate
By definition, a mechanical wave requires a physical medium (solid, liquid, or gas) to travel. This is because mechanical waves rely on the vibration of particles to transport energy from one location to another.
16. 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
Solution
(ii) number of compressions per second Frequency is the number of complete wave cycles (one compression and one rarefaction) that pass a fixed point in one second. If you increase the frequency, the number of compressions (areas of high pressure) passing a point per second must increase.
17. 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
Solution
(iii) .
Frequency Number of compressions = 20 Total time Frequency
18. 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 sound will produce reverberation, not a distinct echo. The human ear can only distinguish two sounds as separate (an echo) if they reach the ear at least 0.1 seconds apart. If a reflected sound arrives in less than 0.1 seconds, our brain "mixes" the original sound and the reflection together. Instead of hearing a repeat, we simply hear the original sound lasting a bit longer. In our case, the time interval is 0.05 seconds. Since , the reflection arrives too quickly to be heard as a distinct echo.
19. Graphs representing two sound waves are given in figure. If the scales on the X and axes of the two graphs are the same, which of the two sound waves has (i) greater wavelength, and (ii) smaller amplitude?
Solution
Wave (i) has the greater wavelength. Wavelength is the distance between two consecutive peaks (compressions) or two consecutive depths (rarefactions). In the provided graphs, the horizontal distance between the peaks in the first graph is visibly longer than the distance between the peaks in the second graph. Since the scales on the X-axis (Distance) are the same, a wider gap indicates a larger wavelength.
Wave (ii) has the smaller amplitude. Amplitude is represented by the maximum displacement or the height of the peak from the mean (center) line on the Y-axis (Density). Comparing the vertical height of the " compressions " or the depth of the " rarefactions," the peaks in the second graph do not rise as high as those in the first graph. Because the Y-axis scales are identical, the shorter peak corresponds to a lower maximum change in density, and thus a smaller amplitude.
20. The sound waves emitted by three sources and C are represented in Fig. 10.31. If the frequency of A is maximum and is minimum, identify the corresponding curves, and mark and C on them.
Solution
Source A (Maximum Frequency): This curve must have the most cycles in a given time. The green curve oscillates the fastest.
Source C (Minimum Frequency): This curve must have the fewest cycles in a given time, stretched out over the distance. The blue curve has the longest wavelength and lowest frequency.
Source B (Intermediate Frequency): The red curve falls between the green and blue in terms of how often it repeats.
21. Draw a graph to represent a sound wave for which the density amplitude is 3 units and wavelength is 4 cm .
Solution
Solution
The biggest error is the simultaneous arrival of light and sound. Light and sound travel at vastly different speeds through the Earth's atmosphere:
- Speed of Light (c): Approximately 300,000,000 meters per second.
- Speed of Sound (v): Approximately 343 meters per second (at sea level, ). Because light is roughly 874,000 times faster than sound, you would see the flash almost instantly, regardless of how far away the bomb is. The sound, however, would take about 3 seconds to travel every kilometer (or about 5 seconds per mile).
23. A source produces a sound wave of wavelength 3.44 m . If the wave travels with a speed of find its time period.
Solution
Given that: Wavelength Speed
24. A ship searching for a sunken ship sent a sonar signal and detected an echo after 5 s . If ultrasonic wave travels at in seawater, approximately how far down in the ocean is the wreckage of the sunken ship located?
Solution
Given Data: Speed of ultrasonic wave (v): Time for echo ( t ): 5 s The total distance traveled by the wave is: Total distance Speed × Time
25. 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 .
Solution
Distance to the obstacle (d) = 1.2 m Total distance traveled by the wave (2d) times 1.2 m Speed of the ultrasonic wave ( v ) Time ( milliseconds)
26. The speed of sound in air is about at and nearly at . Roughly how much extra time will the sound of thunder take to travel a distance of 1720 m , if the air temperature changes from to ? Assume that all other conditions remain unchanged.
Solution
Distance Speed of sound at Speed of sound at time taken by the sound to travel the distance at time taken by the sound to travel the distance at approx the total extra time required (approx)
27. The variation of density of medium for a sound wave propagating with a speed of is shown in Fig. 10.32. Calculate the wavelength and frequency of the sound wave.
Solution
28. The graphical representation of two sound waves A and B propagating at the same speed of is shown in figure. What is the wavelength of each of them? Also, calculate their frequencies.
Solution
For Wave A, one full cycle (one compression and one rarefaction) is completed at 5 cm .
For Wave B, one full cycle is completed at 2.5 cm .
Calculations for Wave A Wavelenght Frequency Calculations for Wave B Wavelength Frequency (f)
29. Two identical sound sources are placed at and one in air and one submerged in water figure. 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
Let the distance from each sound source ( and ) to the vertical cliff be . Since the sound travels to the cliff and reflects back to its source, the total distance traveled by the sound waves in both cases is 2d. the speed of sound in air (Source A). the speed of sound in water (Source B). the time taken for the sound to return to A . the time taken for the sound to return to B . We are given that the time taken by the sound to return to A is 4.5 times that of B :
4.0Key Topics in NCERT Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications
5.0Mind Map / Concept Recap
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