Class 9 Science Chapter 10: Sound Waves — Solutions

These Class 9 Science Exploration Chapter 10 solutions cover Sound Waves: Characteristics and Applications from NCERT’s Exploration, first edition April 2026 (printed pages 184–207).

Class: 9 Subject: Science Book: Exploration Chapter: 10 Exercise: Revise, Reflect, Refine (15 Qs) Session: 2026–27
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Class 9 Science Exploration Chapter 10 Solutions – Overview

Chapter 10 of Exploration, Sound Waves: Characteristics and Applications, explains how sound is produced by vibrating bodies and travels as a longitudinal mechanical wave through a medium as a series of compressions and rarefactions. It covers the characteristics of a sound wave — wavelength, frequency, time period, amplitude and speed — the key relation v = f × λ, the ideas of echo and reverberation, and important applications such as SONAR and ultrasound. The worked solutions include scaled density graphs, echo travel paths and calculations with consistent units.

Key Concepts & Definitions

Sound in air is a longitudinal mechanical wave — it needs a medium and the particles vibrate along the direction of travel.

Compression: a high-density, high-pressure region; rarefaction: a low-density, low-pressure region.

Wavelength (λ): distance between two consecutive compressions (or rarefactions). Frequency (f): number of waves passing a point per second (Hz). Time period (T): time for one wave = 1f\frac1f. Amplitude: the maximum change in density/pressure.

Echo: a distinct reflected sound heard when the gap is at least 0.1 s. Reverberation: persistence of sound by repeated reflections when the gap is less than 0.1 s.

SONAR & ultrasound: sonar uses sound echoes (often ultrasound) for ranging; ultrasound means sound above 20 kHz, which also has imaging and other applications.

Sound Wave Formulas

v=fλ,f=1T,T=1fv=f\lambda,\qquad f=\frac1T,\qquad T=\frac1ff=NΔtf=\frac{N}{\Delta t}

For a stationary source/receiver and reflector with a shared outward/return path:

2d=vt,d=vt22d=vt,\qquad d=\frac{vt}{2}

The roughly 0.1 s separation used in school echo problems is an approximate hearing criterion, not a universal sharp boundary. Reverberation means persistence from repeated reflections; audibility also depends on source duration and reflected intensity.

“Think It Over” — Answers

Two astronauts standing close together in space cannot hear each other speak directly. Why?

ANSWERSound is a mechanical wave and needs a material medium to travel. Space is almost a vacuum with no air, so the sound cannot propagate — the astronauts must use radios, which work with electromagnetic waves that can travel through vacuum.

How is sound produced, and how does it reach our ears?

ANSWERSound is produced by a vibrating object. The vibrations push and pull the surrounding air, creating compressions and rarefactions that travel out as a longitudinal wave; on reaching our ears they make the eardrum vibrate, and we hear the sound.

Textbook Activities and Pause and Ponder

Think It Over: Astronauts and bats

Astronauts use radios to communicate across the near-vacuum; ordinary airborne speech and clanking do not cross that vacuum. Mechanical vibrations can travel along structures in contact. Most echolocating bats emit short high-frequency calls and detect reflected sound to locate prey.

Activities 10.1–10.8: What to observe

  1. A plucked rubber band vibrates. Changing tension, length and thickness can change frequency; compare only one factor at a time. A box can couple vibrations to more air and make the sound easier to hear.
  2. A gently struck tuning fork vibrates; touching water reveals ripples/splashes. Avoid striking hard surfaces or bringing a vibrating prong into contact with the ear.
  3. A desk can carry sound through a solid as well as through surrounding air.
  4. Tapping submerged spoons demonstrates that water transmits sound; real listening paths may also include the vessel and air.
  5. The slinky disturbance travels while a marked coil oscillates about its mean position. Longitudinal vibrations are parallel to propagation.
  6. Sound can transfer energy to the stretched membrane and grains. Mechanical contact or table vibration can confound the observation, so keep those paths separate.
  7. Measure actual musical-note frequencies and compare ratios to Sa. An octave has approximately double frequency; do not invent readings or assume every musical scale has equally spaced frequencies.
  8. At a comfortable level, increasing frequency usually raises perceived pitch. Hearing range varies among listeners; phone speakers and apps have limits and cannot diagnose hearing.

Pause and Ponder: Answers 1–9

  1. Samples: plucking strings, striking a membrane or metal object, and blowing across an air column; each starts vibrations.
  2. Sitar/guitar: strings; tabla/drum: membrane; flute: air column; cymbals: metal bodies. Some instruments couple several vibrating parts.
  3. Option (ii): A and R true, and R explains A in the bell-jar model. In a real apparatus some structure-borne sound may remain.
  4. Option (iii): the disturbance travels but particles oscillate; they do not continuously travel with the wave.
  5. Option (ii): energy carried by the wave.
  6. Mark dense regions C and sparse regions R; graph density on the vertical axis and distance on the horizontal. Density maxima align with compression centres, minima with rarefaction centres.
  7. With comparable material, length and tension, a thinner band generally has lower mass per length and higher frequency, hence shorter period. Report observations if tension also changes.
  8. 20 oscillations each second means 20 × 60 = 1200 oscillations per minute.
  9. In Fig. 10.19, one complete repeat spans 3.0 cm, so half a wavelength is 1.5 cm.

Pause and Ponder: Answers 10–13

(10) Using the supplied table:

vwatervair=1500340=7517≈4.41\frac{v_{\mathrm{water}}}{v_{\mathrm{air}}}=\frac{1500}{340}=\frac{75}{17}\approx4.41vsteelvwater=50001500=103≈3.33\frac{v_{\mathrm{steel}}}{v_{\mathrm{water}}}=\frac{5000}{1500}=\frac{10}3\approx3.33

(11) For a 340 m path:

tair=340340 s=1.00 st_{\mathrm{air}}=\frac{340}{340}\,\mathrm s=1.00\,\mathrm ststeel=3405000 s=0.068 st_{\mathrm{steel}}=\frac{340}{5000}\,\mathrm s=0.068\,\mathrm sΔt=0.932 s\Delta t=0.932\,\mathrm s

The steel signal arrives first. The separation exceeds the textbook’s approximate 0.1 s criterion, so two sounds can be distinguished if both are sufficiently audible.

(12) Minimum reflector distance:

dmin⁡=343(0.2)2 m=34.3 md_{\min}=\frac{343(0.2)}{2}\,\mathrm m=34.3\,\mathrm m

(13) Sonar depth:

d=1500(4)2 m=3000 md=\frac{1500(4)}2\,\mathrm m=3000\,\mathrm m

Journey Beyond: Sample methods

Compare a paper cone with and without the same sound source at fixed positions and volume; a cone redirects and couples sound rather than creating energy. Curved reflectors can distribute sound but may also focus it and create uneven loudness; combine reflection control with absorption and diffusion. For the open-ground speed experiment, use actual measured distance and repeated light-to-sound delays:

v≈dtˉv\approx\frac d{\bar t}

Seeing the burst marks emission almost immediately; manual reaction time and distance measurement can be major errors. Do not present sample numbers as measured results. The hearing-health article should use current authoritative research and local scheme eligibility; an app test is not a clinical hearing assessment.

Class 9 Science Exploration Chapter 10 Solutions — Revise, Reflect, Refine

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

ANSWERCorrect option: (ii) Sound needs a medium to propagate. A mechanical wave can travel only through a material medium — this is what distinguishes it from electromagnetic waves, which travel through vacuum.

2. For a sound wave in a medium, increasing its frequency will increase its (i) wavelength (ii) speed (iii) number of compressions per second (iv) time period

ANSWERCorrect option: (iii) number of compressions per second. Frequency is the number of compressions passing per second; in the same medium the speed stays the same, so a higher frequency means a shorter wavelength and a smaller time period.

3. 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

ANSWER

Option (ii):

f=NΔt=204 s=5 Hzf=\frac N{\Delta t}=\frac{20}{4\,\mathrm s}=5\,\mathrm{Hz}

4. In a room, the reflected sound reaches the ear 0.05 s after its production. Will it produce an echo or reverberation? Justify.

ANSWER An echo can be heard only if the reflected sound reaches the ear at least 0.1 s after the original sound. Here the gap is only 0.05 s, which is less than 0.1 s, so the reflected sound merges with the original — it produces reverberation, not an echo.

5. Graphs of two sound waves are given in Fig. 10.30 (same scales on both axes). Which wave has (i) greater wavelength, and (ii) smaller amplitude?

ANSWER (i) Wave (a) has the greater wavelength — it shows fewer waves (compressions are farther apart) over the same distance. (ii) Wave (a) also has the smaller amplitude — its density variation (height of the curve about the mean) is smaller than that of wave (b). So(a) is the lower-frequency, smaller-amplitude wave, while(b) is the higher-frequency, larger-amplitude wave.

6. The sound waves from three sources A, B and C are shown in Fig. 10.31. If the frequency of A is maximum and that of C is minimum, identify the curves and mark A, B and C.

ANSWER Frequency depends on the number of waves in a given distance (a shorter wavelength means a higher frequency), not on amplitude. A = the curve with the most compressions (shortest wavelength) — the green curve. C = the curve with the fewest compressions (longest wavelength) — the blue curve. B = the curve in between — the red curve.

7. Draw a graph of a sound wave with density amplitude 3 units and wavelength 4 cm.

ANSWER

Plot density change from its mean against distance. One suitable curve is:

Δρ=3sin⁡(2πx4 cm)\Delta\rho=3\sin\left(\frac{2\pi x}{4\,\mathrm{cm}}\right)
Density change versus distance: peaks plus 3 units and troughs minus 3 units, with successive peaks at 1 and 5 cm giving wavelength 4 cm.
An original scaled graph. Negative density changes mean below-average density, not negative absolute density. The plotted sinusoid does not mean air particles move transversely. Open full-size diagram.

8. In a movie, the explosion of a spacecraft in space is shown with a flash of light and sound at the same time. What are the errors in this depiction?

ANSWER

A distant observer cannot hear an explosion through a vacuum because ordinary sound needs a medium; light can reach the observer. If both signals travel through a material medium over an appreciable distance, the light arrives much earlier. Near a source the delay can be too small to notice. Sound conducted through a spacecraft’s structure or heard inside its air-filled cabin is a different situation from sound crossing empty space.

9. A source produces a sound wave of wavelength 3.44 m travelling at 344 m s-1. Find its time period.

ANSWERf=vλ=344 m s−13.44 m=100 Hzf=\frac v\lambda=\frac{344\,\mathrm{m\,s^{-1}}}{3.44\,\mathrm m}=100\,\mathrm{Hz}T=1f=1100 Hz=0.010 sT=\frac1f=\frac1{100\,\mathrm{Hz}}=0.010\,\mathrm s

10. A ship sends a sonar signal and detects an echo after 5 s. If the ultrasonic wave travels at 1525 m s-1 in seawater, how far down is the sunken ship?

ANSWER

For a target directly below the ship, the measured 5 s is the two-way travel time.

Ship sonar sends sound to the seabed and receives an echo; each one-way path has depth d and the total path is 2d.
Sonar travel-path schematic, not to scale. For an oblique target, the same calculation gives slant range rather than vertical depth. Open full-size diagram.
d=vt2=(1525 m s−1)(5 s)2d=\frac{vt}{2}=\frac{(1525\,\mathrm{m\,s^{-1}})(5\,\mathrm s)}{2}d=3812.5 m≈3.8 kmd=3812.5\,\mathrm m\approx3.8\,\mathrm{km}

11. A parking sensor emits an ultrasonic wave that reflects off an obstacle. When the beep starts at 1.2 m from the obstacle, how much time does the wave take to travel to the obstacle and back? (Speed in air = 345 m s-1.)

ANSWERt=2dv=2(1.2 m)345 m s−1t=\frac{2d}{v}=\frac{2(1.2\,\mathrm m)}{345\,\mathrm{m\,s^{-1}}}t≈0.00696 s=6.96 mst\approx0.00696\,\mathrm s=6.96\,\mathrm{ms}

This is the acoustic round trip; an electronic warning system may add processing delay.

12. The speed of sound is about 331 m s-1 at 0 °C and 344 m s-1 at 22 °C. Roughly how much extra time will the sound of thunder take to travel 1720 m if the temperature changes from 22 °C to 0 °C?

ANSWERt22=1720344 s=5.00 st_{22}=\frac{1720}{344}\,\mathrm s=5.00\,\mathrm st0=1720331 s≈5.196 st_0=\frac{1720}{331}\,\mathrm s\approx5.196\,\mathrm sΔt=t0−t22≈0.196 s≈0.20 s\Delta t=t_0-t_{22}\approx0.196\,\mathrm s\approx0.20\,\mathrm s

Sound takes longer in the colder air under the question’s unchanged-other-conditions assumption.

13. The density variation of a sound wave travelling at 340 m s-1 is shown in Fig. 10.32. Calculate the wavelength and frequency.

ANSWER

The 8 cm arrow spans two compression-to-compression intervals (three compression centres), hence two wavelengths:

λ=8 cm2=4 cm=0.040 m\lambda=\frac{8\,\mathrm{cm}}2=4\,\mathrm{cm}=0.040\,\mathrm mf=340 m s−10.040 m=8500 Hzf=\frac{340\,\mathrm{m\,s^{-1}}}{0.040\,\mathrm m}=8500\,\mathrm{Hz}

14. Two sound waves A and B (Fig. 10.33) travel at the same speed of 345 m s-1. Find the wavelength and frequency of each.

ANSWER

Read equal-phase points on the same distance axis. Wave A repeats every 2.5 cm, while B repeats every 5.0 cm:

λA=0.025 m,λB=0.050 m\lambda_A=0.025\,\mathrm m,\qquad\lambda_B=0.050\,\mathrm mfA=3450.025 Hz=13 800 Hzf_A=\frac{345}{0.025}\,\mathrm{Hz}=13\,800\,\mathrm{Hz}fB=3450.050 Hz=6900 Hzf_B=\frac{345}{0.050}\,\mathrm{Hz}=6900\,\mathrm{Hz}

15. Two identical sound sources at A (in air) and B (in water) send sound horizontally to a cliff and back (Fig. 10.34). If the time taken to return to A is 4.5 times that to B, what is the ratio of the speeds of sound in air and water?

ANSWER

Both sounds travel the same horizontal round-trip distance, so speed is inversely proportional to travel time:

vairvwater=tBtA\frac{v_{\mathrm{air}}}{v_{\mathrm{water}}}=\frac{t_B}{t_A}vairvwater=14.5=29\frac{v_{\mathrm{air}}}{v_{\mathrm{water}}}=\frac1{4.5}=\frac29

Speed in air : speed in water = 2 : 9.

Common Mistakes to Avoid

Watch out for these

  • Thinking sound can travel through vacuum — it cannot; it needs a medium.
  • Mixing up frequency (waves per second) with wavelength (distance per wave) — they are inversely related at fixed speed.
  • Forgetting the factor of 2 in echo/SONAR problems (the sound travels to the object and back).
  • Not converting cm to m before using v = f × λ.
  • Confusing echo (gap ≥ 0.1 s, distinct repeat) with reverberation (gap < 0.1 s, persistence).
  • Assuming amplitude affects frequency — loudness depends on amplitude, pitch depends on frequency.

Extra Practice Questions

Very Short Answer Type Questions

Q1. What type of wave is a sound wave?

ANSWERA longitudinal mechanical wave.

Q2. What is the SI unit of frequency?

ANSWERThe hertz (Hz).

Q3. On which property of a sound wave does its loudness depend?

ANSWERIts amplitude.

Short Answer Type Questions

Q1. Distinguish between the pitch and the loudness of a sound.

ANSWERPitch depends on the frequency — a higher frequency gives a higher pitch (shriller sound); loudness depends on the amplitude — a larger amplitude gives a louder sound.

Q2. A sound wave has a frequency of 500 Hz and a wavelength of 0.66 m. Calculate its speed.

ANSWERv = f × λ = 500 × 0.66 = 330 m s-1.

Long Answer Type Question

Q1. Explain how sound propagates through air using the idea of compressions and rarefactions.

ANSWER When an object vibrates, it pushes the nearby air particles together, forming a region of high density and pressure called a compression. As the object moves back, the air particles spread apart, forming a region of low density and pressure called a rarefaction. These compressions and rarefactions move forward one after another, carrying the sound energy through the air as a longitudinal wave, while the air particles only vibrate to and fro about their mean positions.

MCQs & Assertion–Reason

1. Sound waves are:

(a) transverse    (b) longitudinal    (c) electromagnetic    (d) stationary

2. Sound cannot travel through:

(a) solids    (b) liquids    (c) gases    (d) vacuum

3. The SI unit of frequency is:

(a) metre    (b) second    (c) hertz    (d) decibel

4. The relation between speed, frequency and wavelength is:

(a) v = f / λ    (b) v = f × λ    (c) v = λ / f    (d) v = f + λ

5. The minimum time gap to hear a distinct echo is:

(a) 0.01 s    (b) 0.1 s    (c) 1 s    (d) 10 s

6. The pitch of a sound depends on its:

(a) amplitude    (b) frequency    (c) speed    (d) wavelength only

7. SONAR is used to measure:

(a) air temperature    (b) depth of the sea    (c) light intensity    (d) wind speed

8. The frequency of ultrasound is:

(a) below 20 Hz    (b) 20 Hz to 20 000 Hz    (c) above 20 000 Hz    (d) exactly 20 Hz

9. If a sound wave has frequency 200 Hz, its time period is:

(a) 0.005 s    (b) 0.05 s    (c) 200 s    (d) 5 s

10. The loudness of a sound depends on its:

(a) frequency    (b) wavelength    (c) amplitude    (d) speed

Answer key: 1-(b), 2-(d), 3-(c), 4-(b), 5-(b), 6-(b), 7-(b), 8-(c), 9-(a), 10-(c).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: Sound cannot travel in a vacuum.

Reason: Sound is a mechanical wave that needs a medium to propagate.

A-R 2. Assertion: An echo is heard only if the reflecting surface is far enough away.

Reason: The reflected sound must reach the ear at least 0.1 s after the original.

A-R 3. Assertion: A higher-frequency sound has a higher pitch.

Reason: Pitch is determined by the amplitude of the wave.

A-R 4. Assertion: Sound travels faster in water than in air.

Reason: Speed depends on both elastic response and density; their combined effect gives a higher sound speed in water under the stated ordinary conditions.

A-R 5. Assertion: SONAR uses ultrasound to find the depth of the sea.

Reason: Ultrasound is reflected by the seabed and the echo time is measured.

Answer key: 1-(A), 2-(A), 3-(C), 4-(A), 5-(A).

Quick Revision Summary

  • Sound is a longitudinal mechanical wave; it needs a medium and travels as compressions and rarefactions.
  • v = f × λ and T=1fT=\frac1f.
  • Pitch depends on frequency; loudness depends on amplitude.
  • Echo: gap ≥ 0.1 s (distinct); reverberation: gap < 0.1 s (persistence).
  • In echo/SONAR problems use 2d = v × t (sound goes and returns).
  • Sound travels faster in solids and liquids than in gases.

Real-life Applications

Sound science is everywhere: SONAR measures sea depth and locates shipwrecks and shoals of fish; ultrasound scans are used in medicine and to clean and detect flaws in metals; bats and dolphins use echolocation to navigate; parking sensors use ultrasonic echoes; and concert halls are designed to control reverberation for clear sound.

How to revise this chapter

Always write v = f × λ with consistent units (convert cm to m), and remember the factor of 2 for echoes and SONAR. Keep pitch (frequency) and loudness (amplitude) separate, and quote the 0.1 s rule when distinguishing an echo from reverberation.

Frequently Asked Questions

What is Class 9 Science Exploration Chapter 10 about?

Sound waves — how sound is produced and travels as a longitudinal wave, its characteristics (wavelength, frequency, amplitude, speed), v = f × λ, echo and reverberation, and applications such as SONAR and ultrasound.

What is the relation v = f × λ?

The speed of a wave equals its frequency multiplied by its wavelength; the time period is the reciprocal of the frequency (T=1fT=\frac1f).

What is the difference between an echo and reverberation?

An echo is a distinct repeat heard when the reflected sound arrives at least 0.1 s later; reverberation is the persistence of sound from repeated reflections when the gap is shorter than 0.1 s.

Are these Class 9 Science Exploration Chapter 10 solutions free?

Yes. All solutions are free and follow the official NCERT Exploration textbook for 2026–27.

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