These questions appear inside the chapter itself, not at the end. The book page number is shown against each one so you can find it while reading.
Explore various ways of producing sound.
Every sound is produced by something vibrating — moving rapidly to and fro. So the different ways of producing sound are really different ways of setting an object (or the air) into vibration:
In each case, when the vibration stops, the sound stops too — which is why we say sound is produced by vibrating objects.
Make a list of different types of musical instruments and identify their vibrating parts which produce sound.
Musical instruments can be grouped by the part that vibrates:
| Instrument | Part that vibrates |
|---|---|
| Sitar, veena, tanpura, guitar | Stretched strings (plucked) |
| Sarangi, violin | Stretched strings (bowed) |
| Tabla, mridangam, dholak, drum | Stretched membrane (skin) of the drum head |
| Bansuri (flute), shehnai, whistle | Air column inside the pipe |
| Harmonium, mouth organ | Thin metal reeds, with the air around them |
| Ghanta (bell), manjira, jal tarang | The metal (or the bowl and its water) when struck |
| Ghatam (clay pot), triangle | The body of the pot / the metal rod when struck |
In most instruments more than one part vibrates — for example, in a sitar the strings vibrate and the hollow wooden body (the gourd) vibrates with them, making the sound louder and richer.
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:
Correct option: (ii) — both A and R are true, and R is the correct explanation of A.
A is true. This is the vacuum bell-jar experiment: as the air is pumped out, the ringing becomes fainter and almost no sound is heard once a near vacuum is reached, even though the bell can still be seen ringing.
R is true. Sound is a mechanical wave — it travels as compressions and rarefactions of the particles of a medium (solid, liquid or gas). With no particles, there is nothing to carry the vibrations.
R explains A. We cannot hear the bell precisely because the medium (air) between the bell and our ear has been removed, and sound cannot travel without a medium.
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:
Correct option: (iii) — A is true, but R is false.
A is true. When a source vibrates, regions of higher density (compressions) and lower density (rarefactions) are formed and these regions travel away from the source through the medium.
R is false. The particles of the medium do not travel with the wave. Each particle only vibrates back and forth about its own mean position, parallel to the direction of propagation, and passes the disturbance on to its neighbours by collisions. Remember the slinky: the coloured mark on one turn only moves back and forth, while the crowded and spread-out regions travel along the slinky.
What travels is the disturbance and the energy, not the particles.
When sound travels from a tuning fork to your ear, which of the following actually reaches your ear?
Correct option: (ii) — energy carried by sound waves.
The prongs of the tuning fork push and pull on the air next to them, creating compressions and rarefactions. These pass from layer to layer of air by collisions, but each air particle only vibrates about its own position. So:
Only the energy of the vibration is passed on and reaches the eardrum, making it vibrate.
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).
Labelling C and R. Where the dots are crowded, the density is above average — that is a compression (C). Where the dots are spread out, the density is below average — a rarefaction (R). Reading along each strip at the dashed lines:
Labelling the axes. In graphs (c) and (d) the horizontal (x) axis is Distance and the vertical (y) axis is Density of the medium. The dashed horizontal line is the average density.
Drawing the curves. Under every C the density is maximum, so the curve has a crest; under every R it has a trough; midway between them the curve crosses the average-density line. One crest to the next crest takes 4 of the 10 equal divisions, so both waves have the same wavelength:
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?
Yes — stretched to the same tension, the thin rubber band vibrates faster (it completes more vibrations every second) than the thick one, which is heavier.
Frequency. Frequency is the number of oscillations per second. Since the thin band makes more vibrations each second, the sound it produces has a higher frequency (we hear it as a higher-pitched, shriller sound). The thick band gives a lower frequency, deeper sound.
Time period. The time period is the time for one oscillation, and it is related to frequency by
So a higher frequency means a shorter time period. The sound from the thin band has a higher frequency and a smaller time period than the sound from the thick band.
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?
A frequency of 20 Hz means the piston completes 20 oscillations every second. One minute is 60 seconds, so
The piston completes 1200 oscillations per minute.
For the sound wave represented by the graph shown in Fig. 10.19, what is half of its wavelength?
Read one complete wave from the graph. The curve starts on the average-density line at 0, rises to a crest at 0.75 cm, comes back to the average line at 1.5 cm, falls to a trough at 2.25 cm and returns to the average line, going up again, at 3.0 cm. The pattern then repeats (the next crest is at 3.75 cm, which is 3.0 cm after the first crest at 0.75 cm).
The wavelength is the distance between two consecutive crests:
Half of the wavelength is 1.5 cm. On the graph this is the distance from a crest to the next trough (a compression to the next rarefaction), for example from 0.75 cm to 2.25 cm.
Table 10.1 shows the speed of sound in a few media at atmospheric pressure.
Table 10.1: Speed of sound in different media at 15 °C
| State | Medium | Approx. speed |
|---|---|---|
| Solid | Steel | 5000 m s−1 |
| Liquid | Water | 1500 m s−1 |
| Gas | Air | 340 m s−1 |
Compare the speeds in different media by finding the ratio of
(i) Water with respect to air.
Sound travels about 4.4 times faster in water than in air (ratio 75 : 17).
(ii) Steel with respect to water.
Sound travels about 3.3 times faster in steel than in water (ratio 10 : 3). The units cancel in each ratio, so the answers are pure numbers. Together they show the rule from the chapter: sound is fastest in solids, slower in liquids and slowest in gases.
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.)
Given: distance d = 340 m; speed in air = 340 m s−1; speed in steel = 5000 m s−1. Time = distance ÷ speed.
Time through air:
Time through steel:
Time difference:
The sound through the steel reaches Gunjan first, and the sound through the air arrives about 0.93 s later. Since 0.932 s is much more than 0.1 s, yes — she can hear the two sounds separately: first a knock through the fence, then the same knock through the air.
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.
For an echo, the sound travels to the surface and back, so in time t it covers twice the distance d of the surface:
The reflecting surface must be placed at least 34.3 m away. (Any nearer and the echo would return in less than 0.2 s.)
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?
The sonar signal goes down to the sea bed and comes back up, so the 4 s is the time for twice the depth. The time to reach the sea bed is half of it:
The ocean is 3000 m (3 km) deep at that location.
Which observation best supports the idea that sound is a mechanical wave?
Correct option: (ii) — sound needs a medium to propagate.
A mechanical wave is, by definition, a wave that needs a material medium for its propagation. The vacuum bell-jar experiment shows that sound cannot travel without a medium, so this observation is exactly what makes sound a mechanical wave.
The other options do not separate mechanical waves from other waves: light also shows reflection, has a frequency and carries energy — yet light travels through vacuum and is not a mechanical wave.
For a sound wave propagating in a medium, increasing its frequency will increase its
Correct option: (iii) — number of compressions per second.
Frequency is the number of density oscillations at a point per second; each oscillation brings one compression. So a higher frequency means more compressions pass a point every second.
If 20 compressions pass a point in 4 seconds, the frequency is
Each compression passing a point marks one complete oscillation, so 20 oscillations take place in 4 s.
Correct option: (ii) 5 Hz. (Option (i) wrongly multiplies, and option (iv), 0.2 s, is the time period \(T = 1/\nu\), not the frequency.)
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.
It will produce reverberation, not an echo.
To hear an echo as a separate sound, the reflected sound must reach us 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 brain cannot separate the reflected sound from the original — no distinct echo is heard.
Instead, such quickly returning reflections (the book gives the condition as a time difference of about 0.05 s or less) overlap with and prolong the original sound, so the sound seems to persist after the source stops. This persistence due to multiple reflections in a room is reverberation.
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?
Since both graphs are drawn on the same scales, we can compare them directly.
So wave (a) has both the greater wavelength and the smaller amplitude — it is a lower-pitched and softer sound than (b).
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.
All three waves travel in the same medium (air), so they have the same speed. From \(v = \lambda \times \nu\), the wave with the shortest wavelength (the most waves packed into the same distance) has the highest frequency. Count the complete waves (crest to crest) across the graph:
| Curve (complete waves) | Source |
|---|---|
| Green — 4 waves (shortest \(\lambda\)) | A (maximum frequency) |
| Red — 3 waves | B |
| Blue — 2 waves (longest \(\lambda\)) | C (minimum frequency) |
Draw a graph to represent a sound wave for which the density amplitude is 3 units and wavelength is 4 cm.
How to draw it:
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?
There are two errors:
So showing the flash and the sound at the same moment is wrong on both counts.
A source produces a sound wave of wavelength 3.44 m. If the wave travels with a speed of 344 m s−1 find its time period.
In one time period the wave travels exactly one wavelength, so \(v = \lambda / T\), which gives
The time period is 0.01 s. (Its frequency is therefore 1 / 0.01 s = 100 Hz.)
A ship searching for a sunken ship sent a sonar signal and detected an echo after 5 s. If ultrasonic wave travels at 1525 m s−1 in seawater, approximately how far down in the ocean is the wreckage of the sunken ship located?
The 5 s is the time for the signal to go down to the wreckage and come back, so the distance d satisfies 2d = v × t:
The wreckage is about 3812.5 m, i.e. roughly 3.8 km, below the ship.
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 m s−1.
The wave travels to the obstacle and back, a total distance of 2 × 1.2 m = 2.4 m.
The round trip takes about 0.007 s (7 milliseconds). The 40 kHz frequency is not needed for this calculation — it only tells us the wave is ultrasonic (above 20 kHz), so the driver does not hear it.
The speed of sound in air is about 331 m s−1 at 0 °C and nearly 344 m s−1 at 22 °C. Roughly how much extra time will the sound of thunder take to travel a distance of 1720 m, if the air temperature changes from 22 °C to 0 °C? Assume that all other conditions remain unchanged.
Time at 22 °C:
Time at 0 °C:
Extra time:
At 0 °C the thunder takes roughly 0.2 s longer (about 0.196 s) to cover 1720 m, because sound travels more slowly in colder air.
The variation of density of medium for a sound wave propagating with a speed of 340 m s−1 is shown in Fig. 10.32. Calculate the wavelength and frequency of the sound wave.
Wavelength. The 8 cm arrow runs from the centre of the first compression (crowded region) to the centre of the third compression. Between them there are two complete waves (compression → compression twice). So
Frequency. Using \(v = \lambda \times \nu\):
Wavelength = 4 cm, frequency = 8500 Hz. Remember to change cm into m before dividing, so that the answer comes out in hertz.
The graphical representation of two sound waves A and B propagating at the same speed of 345 m s−1 is shown in Fig. 10.33. What is the wavelength of each of them? Also, calculate their frequencies.
Reading the wavelengths. Both curves start on the average line at 0.
Frequencies from \(\nu = v / \lambda\):
\(\lambda_A\) = 2.5 cm, \(\nu_A\) = 13800 Hz; \(\lambda_B\) = 5.0 cm, \(\nu_B\) = 6900 Hz. At the same speed, A has half the wavelength of B and so exactly twice its frequency.
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?
A (in air) and B (in water) are at the same horizontal distance d from the vertical face of the cliff. Each sound goes to the cliff and back, a distance 2d.
Dividing the first equation by the second, 2d cancels:
We are told the return time at A is 4.5 times that at B, so
The ratio of the speed of sound in air to that in water is 2 : 9 (sound is 4.5 times faster in water). This matches the book’s statement that sound travels about 4–5 times faster in water than in air.
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