NCERT Solutions Class 9 Science Chapter 10 — Sound Waves: Characteristics and Applications

NCERT Class 9 Science — Sound Waves: Characteristics and Applications, Chapter 10, Revise, Reflect, Refine. All 15 questions solved.

Also here. Solutions to the in-text questions for this chapter and the activities for this chapter are on their own pages.

These are the end-of-chapter questions from Chapter 10 of Exploration, the new Class 9 science textbook NCERT introduced in 2026. They fall into four groups. Questions 1, 2, 3 and 8 are about what sound is: a mechanical wave that needs a medium, described by its frequency. Questions 5, 6, 7, 13 and 14 ask you to read or draw a density–distance graph, where the wavelength is measured along the distance axis and the amplitude up the density axis. Questions 9 and 12 to 14 use the link between speed, wavelength and frequency. Questions 4, 10, 11 and 15 are about reflected sound: echoes, reverberation, sonar and a parking sensor. Everything rests on three results from the chapter:

• Frequency and time period — the frequency $\nu$ is the number of density oscillations at a point in one second, measured in hertz (Hz); the time period $T$ is the time for one oscillation, and $\nu = \dfrac{1}{T}$.
• Speed of sound — a crest moves forward one wavelength in one time period, so $v = \dfrac{\lambda}{T} = \lambda \times \nu$. The speed is set by the medium (and its temperature), not by the source.
• Reflected sound — a reflected sound travels to the surface and back, so the distance to the surface is $\dfrac{v \times t}{2}$. The ear hears the reflection as a separate echo only if it arrives at least $0.1\ \text{s}$ after the original sound.

The book writes frequency as $\nu$ (the Greek letter nu) and speed as $v$. They look alike, so keep track of which is which.

Key insight. In a given medium the speed of sound is fixed, and the source cannot change it. So when the frequency goes up, the wavelength must come down to keep $v = \lambda\nu$ the same (Questions 2 and 6); a wavelength read off a graph gives the frequency at once (Questions 13 and 14); and a measured time turns straight into a distance (Questions 10 and 11). Only a change of medium or of temperature changes the speed, and that change is exactly what Questions 12 and 15 measure.

Question 1

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

Solution. A mechanical wave is a disturbance passed along from particle to particle of a medium, so the feature that marks it out is that it cannot travel without one. Test each option against light, which is a wave but not a mechanical one. Light also reflects, has a frequency and carries energy, so options (i), (iii) and (iv) are true of every kind of wave and cannot show that sound in particular is mechanical. Only option (ii) separates the two: sound needs a medium and cannot cross a vacuum, while light can.

Question 2

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

Solution. Take the four quantities one at a time. The speed of sound is decided by the medium, not by the source, so changing the frequency leaves the speed unchanged; (ii) is ruled out. Since $v = \lambda \times \nu$ and $v$ stays the same, a larger $\nu$ must go with a smaller $\lambda$, so the wavelength decreases; (i) is ruled out. The time period is $T = 1/\nu$, so it decreases too; (iv) is ruled out.

That leaves (iii). Each compression passing a point marks one complete density oscillation there, so the number of compressions passing per second is the frequency itself, and it increases along with the frequency.

Question 3

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

Solution. Each compression that passes the point means one complete density oscillation there, from maximum density to minimum and back. The frequency is the number of oscillations per second, so, as in Example 10.1, divide the number of oscillations by the time taken:

$$\nu = \frac{\text{number of oscillations}}{\text{time taken}} = \frac{20}{4\ \text{s}} = 5\ \text{Hz}$$

Option (iv) is a trap. The time for one oscillation is $T = \dfrac{4\ \text{s}}{20} = 0.2\ \text{s}$; that is the time period, a time in seconds, not the frequency.

Question 4

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. The ear can hear two sounds as separate only if they reach it at least $0.1\ \text{s}$ apart; if the gap is shorter, the brain cannot tell them apart. Here the reflected sound arrives only $0.05\ \text{s}$ after the original, half the time needed, so it cannot be heard as a separate echo. Instead it merges with the original sound, together with the other reflections from the walls, and makes the sound persist for a moment after the source stops. That is reverberation.

Distances tell the same story. Taking the speed of sound in air as $340\ \text{m s}^{-1}$, as the chapter does, the reflected sound travels in $0.05\ \text{s}$ a distance of

$$d = v \times t = 340\ \text{m s}^{-1} \times 0.05\ \text{s} = 17\ \text{m}$$

This is the path to the wall and back, so the wall is only $17 \div 2 = 8.5\ \text{m}$ away, closer than the $17\ \text{m}$ minimum the chapter finds for hearing an echo.

A note on the figure $0.05\ \text{s}$. The chapter describes reverberation as reflections arriving less than $0.05\ \text{s}$ apart, and this question uses exactly $0.05\ \text{s}$, so it sits right at the edge of that description. The deciding test is the echo one: a reflection arriving well short of $0.1\ \text{s}$ cannot be heard as a separate sound, so it adds to the reverberation rather than making an echo.

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?

Solution. Because the two graphs are drawn to the same scales, lengths and heights on them can be compared directly. In the redrawn figure one wavelength is marked on each graph with a grey arrow, and the amplitude with a red one.

(i) Wavelength. The wavelength is the distance between two consecutive crests, measured along the distance axis. Graph (a) fits only three complete waves along its axis, while graph (b) fits six along an axis that is slightly shorter, so each wave of (b) is much shorter. Measured on the figure, one wavelength of (a) is about $2.4$ times one wavelength of (b). Wave (a) has the greater wavelength.

(ii) Amplitude. The amplitude is the maximum change in density from the average, that is, the height of a crest above the dashed average-density line, not above the distance axis. The crests of (b) rise about one and a half times as high above the dashed line as those of (a). Wave (a) has the smaller amplitude.

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.

Solution. The three sounds travel through the same medium, air, so all three waves have the same speed. From $v = \lambda \times \nu$, with $v$ the same for all of them, the wave with the highest frequency must have the shortest wavelength, and the wave with the lowest frequency the longest wavelength.

So count how many complete waves each curve fits into the same distance; all three start together on the left and end together on the right. The green curve fits four complete waves, the red curve three and the blue curve two. The green curve therefore has the shortest wavelength and the highest frequency, which makes it A; the blue curve has the longest wavelength and the lowest frequency, which makes it C; and the red curve, in between, is B. Their frequencies are in the ratio $4 : 3 : 2$.

The heights of the curves play no part in this. Height above the dashed line is amplitude, which decides how loud a sound is, not its frequency; that the green curve is also the tallest is a coincidence of this figure.

Question 7

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

Solution. Draw it the way the chapter draws its density graphs: distance along the horizontal axis, density up the vertical axis, and a dashed horizontal line for the average density of the medium. The amplitude is measured from this dashed line, not from the distance axis, so each crest must be $3$ units above the average line and each trough $3$ units below it. The wavelength is the distance from one crest to the next, so the crests must be $4\ \text{cm}$ apart.

Starting at the average density at distance $0$, the density rises to a crest at $1\ \text{cm}$, falls back to the average at $2\ \text{cm}$, reaches a trough at $3\ \text{cm}$ and returns to the average at $4\ \text{cm}$. That is one complete wave. Repeating it gives the next crest at $5\ \text{cm}$ and the next trough at $7\ \text{cm}$.

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?

Solution. There are two errors. The first is that there should be no sound at all. Sound is a mechanical wave, passed along by the particles of a medium, and space is almost empty, a vacuum. The sound of the explosion has nothing to travel through, so it could never reach an observer. Light needs no medium, so the flash can still be seen.

The second is that even where there is a medium, the flash and the sound would not arrive together. Light travels at about $300\,000\ \text{km s}^{-1}$, while sound in air travels at only about $340\ \text{m s}^{-1}$, so the flash is seen almost at once and the sound arrives noticeably later. This is why lightning is seen before thunder is heard (Example 10.3).

Question 9

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. In one time period the wave moves forward exactly one wavelength, so $v = \dfrac{\lambda}{T}$. Rearranging for the time period:

$$T = \frac{\lambda}{v} = \frac{3.44\ \text{m}}{344\ \text{m s}^{-1}} = 0.01\ \text{s}$$

As a check, go through the frequency instead: $\nu = \dfrac{v}{\lambda} = \dfrac{344}{3.44} = 100\ \text{Hz}$, and $T = \dfrac{1}{\nu} = \dfrac{1}{100} = 0.01\ \text{s}$.

Question 10

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. The $5\ \text{s}$ is the time for the signal to travel down to the wreck and back up to the ship. It reaches the wreck halfway through, after $2.5\ \text{s}$. So, as in Example 10.6, the depth is the speed multiplied by half the time:

$$\text{depth} = \frac{v \times t}{2} = \frac{1525\ \text{m s}^{-1} \times 5\ \text{s}}{2} = \frac{7625}{2}\ \text{m} = 3812.5\ \text{m}$$

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 $40\ \text{kHz}$) 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. The obstacle is $1.2\ \text{m}$ away, and the pulse has to go to it and come back, so it travels twice that distance, $2 \times 1.2 = 2.4\ \text{m}$. Time is distance divided by speed:

$$t = \frac{2 \times 1.2\ \text{m}}{345\ \text{m s}^{-1}} = \frac{2.4}{345}\ \text{s} = 0.00696\ \text{s} \approx 0.007\ \text{s}$$

That is about seven thousandths of a second. The frequency of $40\ \text{kHz}$ tells us the wave is ultrasonic, beyond the $20\ \text{kHz}$ limit of human hearing, but it is not needed for the time, because the speed is given directly.

Question 12

The speed of sound in air is about $331\ \text{m s}^{-1}$ at $0\ ^\circ\text{C}$ and nearly $344\ \text{m s}^{-1}$ at $22\ ^\circ\text{C}$. 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 $22\ ^\circ\text{C}$ to $0\ ^\circ\text{C}$? Assume that all other conditions remain unchanged.

Solution. The distance is the same both times; only the speed changes, because the speed of sound in air depends on its temperature. So find the time taken at each temperature, using time $=$ distance $\div$ speed, and subtract.

At $22\ ^\circ\text{C}$:

$$t_{22} = \frac{1720\ \text{m}}{344\ \text{m s}^{-1}} = 5\ \text{s}$$

At $0\ ^\circ\text{C}$:

$$t_{0} = \frac{1720\ \text{m}}{331\ \text{m s}^{-1}} = 5.196\ \text{s}$$

The extra time is

$$t_{0} – t_{22} = \frac{1720}{331} – 5 = \frac{1720 – 1655}{331} = \frac{65}{331}\ \text{s} \approx 0.2\ \text{s}$$

The colder air is slower, so the thunder arrives about a fifth of a second later.

Question 13

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. In the figure the dots stand for the particles of the medium. Where they are crowded together is a compression (C), where the density is highest; where they are spread out is a rarefaction (R). The wavelength is the distance between the centres of two consecutive compressions. The redrawn figure adds the density graph for the same stretch underneath, to make this easier to see.

The $8\ \text{cm}$ arrow runs from the centre of one compression to the centre of another, and there is a third compression midway between them. So the $8\ \text{cm}$ covers two wavelengths, not one:

$$\lambda = \frac{8\ \text{cm}}{2} = 4\ \text{cm} = 0.04\ \text{m}$$

Now use $v = \lambda \times \nu$, with the wavelength in metres to match the speed in metres per second:

$$\nu = \frac{v}{\lambda} = \frac{340\ \text{m s}^{-1}}{0.04\ \text{m}} = 8500\ \text{Hz}$$

Question 14

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.

Solution. Read each wavelength along the distance axis, as the distance from one crest to the next; equivalently, it is the distance over which the curve goes through one full rise and fall and comes back to the same point of its cycle. The labelled marks on the axis are $2.5\ \text{cm}$ apart.

Wave A. It starts on the average line at $0$, rising, reaches a crest at $0.625\ \text{cm}$, a trough at $1.875\ \text{cm}$, and is back on the average line and rising again at $2.5\ \text{cm}$. Its next crest is at $3.125\ \text{cm}$, exactly $2.5\ \text{cm}$ after the first. So $\lambda_A = 2.5\ \text{cm} = 0.025\ \text{m}$; the graph shows three complete waves of A in $7.5\ \text{cm}$.

Wave B. It also meets the average line at $2.5\ \text{cm}$, but there it is going down, so only half a wave is complete. Its trough is at $3.75\ \text{cm}$, and it is back on the average line and rising at $5.0\ \text{cm}$; its crests are at $1.25\ \text{cm}$ and $6.25\ \text{cm}$, $5.0\ \text{cm}$ apart. So $\lambda_B = 5.0\ \text{cm} = 0.05\ \text{m}$.

Both waves travel at $345\ \text{m s}^{-1}$, so $\nu = \dfrac{v}{\lambda}$ for each:

$$\nu_A = \frac{345\ \text{m s}^{-1}}{0.025\ \text{m}} = 13\,800\ \text{Hz}, \qquad \nu_B = \frac{345\ \text{m s}^{-1}}{0.05\ \text{m}} = 6900\ \text{Hz}$$

Wave A has half the wavelength of B, so at the same speed it has twice the frequency.

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?

Solution. A is directly above B, so both sounds travel the same horizontal distance $d$ to the cliff face and the same distance $d$ back, $2d$ in all. Only the speed differs: A’s sound travels through air and B’s through water. Since time $=$ distance $\div$ speed,

$$t_A = \frac{2d}{v_{\text{air}}}, \qquad t_B = \frac{2d}{v_{\text{water}}}$$

Dividing the first by the second, the $2d$ cancels:

$$\frac{t_A}{t_B} = \frac{v_{\text{water}}}{v_{\text{air}}}$$

For the same distance, the time is inversely proportional to the speed. With $t_A = 4.5\,t_B$,

$$\frac{v_{\text{air}}}{v_{\text{water}}} = \frac{t_B}{t_A} = \frac{1}{4.5} = \frac{2}{9}$$

Sound travels $4.5$ times faster in water than in air, which fits the chapter’s statement that it travels about $4$ to $5$ times faster in water.

Common mistakes

• Question 2: thinking a higher frequency makes sound travel faster. This leads to option (ii). The speed of sound is fixed by the medium; if every pitch travelled at its own speed, the notes of a song played in a large hall would reach the back row out of step. When the frequency rises, the wavelength shrinks instead, so that $v = \lambda\nu$ stays the same.
• Question 3: choosing $0.2\ \text{Hz}$. Dividing $4$ by $20$ gives $0.2\ \text{s}$, the time for one oscillation, which is the time period. The frequency is oscillations per second, $20 \div 4 = 5\ \text{Hz}$. A quick check: one oscillation every $0.2\ \text{s}$ is five in each second.
• Questions 4, 10 and 11: forgetting that reflected sound makes a round trip. In Question 10, not halving gives $7625\ \text{m}$, twice the real depth; in Question 11, using $1.2\ \text{m}$ instead of $2.4\ \text{m}$ gives $0.0035\ \text{s}$, half the real time; and in Question 4 the $17\ \text{m}$ covered in $0.05\ \text{s}$ is to the wall and back, so the wall is only $8.5\ \text{m}$ away. Every echo time is there and back.
• Question 6: matching frequency to the height of the curve. Height above the dashed line is amplitude, which is loudness. Frequency is read from how many waves fit into the same distance: four for green, three for red, two for blue. That the green curve is also the tallest is a coincidence of the figure, not the reason it is A.
• Question 12: dividing the distance by the difference in speeds. $1720 \div (344 – 331) = 132\ \text{s}$ is the time to cover $1720\ \text{m}$ at $13\ \text{m s}^{-1}$, which has nothing to do with the question. Find the time at each speed separately, then subtract the times.
• Question 13: taking the whole $8\ \text{cm}$ as one wavelength. That gives $\nu = 340 \div 0.08 = 4250\ \text{Hz}$. The arrow runs from one compression to the compression after next, so it spans two wavelengths; count the compressions before measuring.
• Question 14: reading $\lambda_B$ as $2.5\ \text{cm}$, or leaving the wavelength in centimetres. Wave B crosses the average line at $2.5\ \text{cm}$, but going downwards; it is only half-way through its cycle there. And $345 \div 2.5 = 138$ is not a frequency in hertz: the wavelength must be in metres, $0.025\ \text{m}$, to match a speed in metres per second.
• Question 15: turning the ratio upside down. For the same distance, the longer time means the lower speed. The sound in air takes $4.5$ times as long, so air is the slower medium and $v_{\text{air}} : v_{\text{water}} = 2 : 9$, not $9 : 2$.

Practise next

• The in-text Pause and Ponder questions of this chapter, thirteen across nine boxes — more stops than any other chapter in the book — running from how a sound is made, through what actually travels, to what sound is used for.
• Chapter 4, Describing Motion Around Us — speed as distance divided by time, the relation behind every echo, sonar and thunder calculation on this page.
• Chapter 7, Work, Energy, and Simple Machines — what energy is and how it is transferred, which is what a sound wave of larger amplitude carries more of.