NCERT Solutions Class 9 Science Chapter 10 In-text Questions — Sound Waves: Characteristics and Applications

NCERT Class 9 Science — Sound Waves: Characteristics and Applications, the nine Pause and Ponder boxes on pages 186 to 203. All 13 questions solved.

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

This chapter stops more often than any other in the book — nine times — and the boxes track it from how sound is made, through what actually travels, to what sound is used for. Four relations carry the numerical questions:

• Frequency and time period — $\nu = \dfrac{\text{number of oscillations}}{\text{time taken}}$, and $T = \dfrac{1}{\nu}$.
• Wave relation — $v = \nu\lambda$, so at a given speed a higher frequency means a shorter wavelength.
• Time to travel — $t = \dfrac{d}{v}$, which is how the same knock arriving through steel and through air is separated in time.
• Echo and sonar — the sound goes out and comes back, so $d = \dfrac{v \times t}{2}$. Forgetting the 2 is the single commonest error in the chapter.

Key insight. In a sound wave the medium does not travel — only the disturbance does. Particles of air oscillate about their own positions and hand the energy on to their neighbours; what reaches your ear from a tuning fork is energy, not the air that was next to the fork. Questions 4 and 5 are that one fact asked twice, and it is also why sound needs a medium at all, which is Question 3.

Pause and Ponder, page 186. The box that follows Section 10.1.1, on the tuning fork.

Question 1

Explore various ways of producing sound.

Solution. The chapter has just shown that a tuning fork makes a sound when its prongs are struck and set vibrating, and that the vibration is real, because a prong touched to water sends ripples across the surface. The point of this question is to notice that every other way of making a sound is the same thing in disguise.

• Plucking — a stretched rubber band, a sitar or guitar string, a veena.
• Striking — a drum or tabla skin, a metal plate, a bell, a steel glass.
• Blowing — a flute, a whistle, a conch: the air column inside is set vibrating.
• Scratching or rubbing — a comb’s teeth, a finger drawn round the rim of a glass.
• Your own voice — air from the lungs sets the vocal cords vibrating.

In every case, something is vibrating. Stop the vibration — put a hand on the bell — and the sound stops at once.

Question 2

Make a list of different types of musical instruments and identify their vibrating parts which produce sound.

Solution. Grouping the instruments by what vibrates is more useful than simply listing them, because it is the same three answers over and over.

Instrument	The part that vibrates
Guitar, sitar, violin, veena, ektara	The stretched string
Tabla, dholak, drum, mridangam	The stretched membrane of the skin
Flute, shehnai, whistle, conch, harmonium	The air column inside the pipe
Cymbals (manjira), triangle, bell, ghatam	The metal or clay body itself
Human voice	The vocal cords

Pause and Ponder, page 188. After Section 10.2, on the propagation of sound.

Question 3

Assertion (A): We cannot hear the sound of 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:

Solution. Test the two statements separately, then ask whether the second explains the first.

Is A true? Yes. This is the bell-jar experiment: as the air is pumped out, the sound grows fainter and fainter, until with most of the air gone almost nothing is heard — although the bell can still be seen to be ringing.

Is R true? Yes. Sound is a mechanical wave. It travels by the particles of a medium being disturbed and passing the disturbance on, so with no particles there is nothing to carry it.

Does R explain A? Exactly. Pumping out the air removes the medium, and removing the medium is precisely why the sound cannot reach the ear. Light needs no medium, which is why the bell remains visible — the contrast that makes the experiment convincing.

Pause and Ponder, page 191. After the section on longitudinal and transverse waves.

Question 4

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:

Solution. A is true. A sound wave is exactly a train of compressions and rarefactions travelling outwards from the source — the chapter’s slinky activity shows the disturbance moving along the spring towards the friend at the other end.

R is false, and this is the misconception the question exists to catch. In that same slinky activity, the mark made on one turn does not travel to the friend. It moves back and forth about its own position while the disturbance passes through. The particles of the medium oscillate about their mean positions; only the disturbance, and the energy it carries, moves forward.

If R were true, air itself would stream out of a loudspeaker and a gale would blow from every ringing bell.

So A is true and R is false: option (iii).

Pause and Ponder, page 192. After Section 10.4, on the energy of sound waves.

Question 5

When sound travels from a tuning fork to your ear, which of the following actually reaches your ear?

Solution. This is the previous question put concretely, and the chapter’s own note answers it: in the propagation of a sound wave it is the energy that is transferred, not the particles of the medium.

Each layer of air near the prong is set oscillating, collides with the layer beyond it, passes on the energy and returns to oscillating about its own position. The disturbance is handed along from neighbour to neighbour all the way to the eardrum, which the arriving energy then sets vibrating.

So (i) is wrong — the air by the fork stays where it is. (iii) is wrong: nothing of the fork leaves it. (iv) is wrong, and is the same error as R in the previous question: there is no stream of air, only an oscillation passed along.

Pause and Ponder, page 193. After Section 10.5, on the graphical representation of a sound wave.

Question 6

The variation of density of the medium for two sound waves is shown in Fig. 10.17 (a) and (b). Label compression and rarefaction by C and R on it. In the graph given in Fig. 10.17 (c) and (d), label the axes and draw the curves corresponding to Fig. 10.17 (a) and (b).

Solution. Work from the dots to the graph, in three steps.

Labelling C and R. A compression is where the particles are crowded together — the dark, dense bands in the picture. A rarefaction is where they are spread out — the pale, sparse bands between them. Mark every dense band C and every sparse band R, so the two alternate all along the strip.

Labelling the axes. The horizontal axis is distance and the vertical axis is density of the medium, with the dashed horizontal line already drawn marking the average density.

Drawing the curve. Line the graph up under the strip so the two share the same distance scale, and then use one rule: the density is highest at the middle of a compression and lowest at the middle of a rarefaction.

• Put a crest (a maximum above the average line) at the centre of each C.
• Put a trough (a minimum below the average line) at the centre of each R.
• Join them into a smooth wave that crosses the average-density line halfway between each crest and trough.

Panels (a) and (b) differ in where their compressions fall, so each curve must follow its own strip: count that panel’s dense bands and put one crest under each.

Pause and Ponder, page 195. After Section 10.6.1, on frequency and time period.

Question 7

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. Stretch the two bands equally and pluck each. The thin band is lighter, so for the same stretch it springs back and forth more times a second: yes, it vibrates faster.

Frequency is the number of oscillations per second, so a faster vibration means a higher frequency. The time period is the time for one oscillation, and

$$T = \frac{1}{\nu}$$

so a higher frequency means a shorter time period. The two always move oppositely.

You hear the difference directly: the thin band gives the shriller, higher-pitched note, and the thick band the deeper one.

Question 8

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. A frequency of 20 Hz means 20 oscillations every second, so the only step is to convert the minute into seconds.

$$1\ \text{minute} = 60\ \text{s}$$

$$\text{oscillations per minute} = 20\ \text{s}^{-1} \times 60\ \text{s} = 1200$$

Question 9

For the sound wave represented by the graph shown in Fig. 10.19, what is half of its wavelength?

Solution. The wavelength is the distance between two successive points in the same phase — most easily read off as the distance between two successive crests of the density curve.

In Fig. 10.19 the first crest is at about $0.75$ cm and the next at about $3.75$ cm, so

$$\lambda = 3.75 – 0.75 = 3.0\ \text{cm}$$

The question asks for half of this:

$$\frac{\lambda}{2} = \frac{3.0}{2} = 1.5\ \text{cm}$$

Half a wavelength is also the distance from a crest to the next trough — from the middle of a compression to the middle of the next rarefaction — which is a useful way to check the reading straight off the graph.

Pause and Ponder, pages 197 and 198. After Section 10.6.3, on the speed of sound. This box runs across the page break — Question 11 is printed on page 198.

Question 10

Table 10.1 shows the speed of sound in a few media at atmospheric pressure. 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. From Table 10.1 at 15 °C: steel $5000\ \text{m s}^{-1}$, water $1500\ \text{m s}^{-1}$, air $340\ \text{m s}^{-1}$.

(i) Water to air:

$$\frac{v_{\text{water}}}{v_{\text{air}}} = \frac{1500}{340} = \frac{75}{17} \approx 4.4$$

(ii) Steel to water:

$$\frac{v_{\text{steel}}}{v_{\text{water}}} = \frac{5000}{1500} = \frac{10}{3} \approx 3.3$$

Both ratios are greater than 1, and the pattern is the point of the question: sound travels fastest in solids, slower in liquids and slowest in gases, because the particles of a solid are held closely and tightly together and pass a disturbance on most quickly.

Question 11

Two friends are standing along a steel fence at a distance of 340 m from each other (Fig. 10.23). 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. One knock sends sound along two routes at once — through the steel of the fence and through the air beside it. Both cover the same 340 m, but at very different speeds, so they arrive at different times. Find each time separately with $t = d/v$.

Through the air, at $340\ \text{m s}^{-1}$:

$$t_{\text{air}} = \frac{340\ \text{m}}{340\ \text{m s}^{-1}} = 1\ \text{s}$$

Through the steel, at $5000\ \text{m s}^{-1}$:

$$t_{\text{steel}} = \frac{340\ \text{m}}{5000\ \text{m s}^{-1}} = 0.068\ \text{s}$$

The sound through the steel arrives first, and the gap between the two is

$$\Delta t = 1 – 0.068 = 0.932\ \text{s}$$

Could she tell them apart? Two sounds are heard separately if they arrive at least 0.1 s apart. Here the gap is 0.932 s, nearly ten times that, so yes — Gunjan would clearly hear two distinct knocks, the steel one first.

Pause and Ponder, page 201. After Section 10.7.1, on echoes.

Question 12

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 343 m s⁻¹.

Solution. The 0.2 s is the time for the round trip — out to the surface and back again. So the sound covers twice the distance to the surface in that time.

$$\text{total path} = v \times t = 343\ \text{m s}^{-1} \times 0.2\ \text{s} = 68.6\ \text{m}$$

The reflecting surface is half of that away:

$$d = \frac{68.6}{2} = 34.3\ \text{m}$$

Any nearer and the echo would return sooner than 0.2 s, so 34.3 m is the minimum.

Pause and Ponder, page 203. The last box, after the section on sonar.

Question 13

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 1500 m s⁻¹?

Solution. Exactly the echo calculation again, with the sea floor as the reflecting surface. The 4 s covers the journey down and back up, so the depth is half the total path.

$$\text{total path} = v \times t = 1500\ \text{m s}^{-1} \times 4\ \text{s} = 6000\ \text{m}$$

$$\text{depth} = \frac{6000}{2} = 3000\ \text{m}$$

Common mistakes

• Forgetting to halve in Questions 12 and 13. The time given is always for the sound to go out and come back, so the distance is $vt/2$. Leaving out the 2 doubles both answers, to 68.6 m and 6000 m.
• Choosing (ii) in Question 4 because both statements sound reasonable. The particles do not travel with the wave — they oscillate about fixed positions, which is why R is false and the answer is (iii).
• Answering Question 5 with (iv), “a stream of compressed air”. Nothing streams. Compressions and rarefactions pass through the air the way a ripple passes along a slinky, while each part of the medium stays where it is.
• Reading the wavelength in Question 9 as crest-to-trough. That distance is half a wavelength. Crest to crest is the full $\lambda = 3.0$ cm, and the question then asks for half of it.
• Giving 20 oscillations per minute in Question 8. A hertz is one oscillation per second; the minute has to be converted, giving 1200.
• Saying in Question 11 that the two sounds cannot be told apart because “sound is fast”. The comparison the question asks for is between the calculated 0.932 s gap and the 0.1 s the ear needs — and 0.932 s is far larger.

Practise next

• The end-of-chapter questions, Revise, Reflect, Refine, where the same wave relation and echo formula appear with harder numbers and in longer word problems.
• Chapter 7, Work, Energy, and Simple Machines — its in-text questions set up the idea of energy being transferred, which is what Questions 4 and 5 here turn on.