Question 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

Accepted Answer

(ii). - (ii) Sound needs a medium to propagate. - A mechanical wave is defined as one that requires a material medium, so this observation alone identifies sound as mechanical. - The bell jar experiment supplies the evidence: the sound fades as air is pumped out and is almost gone at near vacuum, though the bell is still seen ringing. - (i), (iii) and (iv) are all true of sound, but light also reflects, has a frequency and carries energy — and light crosses vacuum, so none of them can prove a wave is mechanical.

Question 2

Explore various ways of producing sound.

Accepted Answer

- Sound is produced whenever something vibrates — every method below is just a different way of setting an object vibrating. - Plucking a stretched band or string — stretch a rubber band across an open cardboard box and pluck it (Activity 10.1); the band is seen vibrating, and the moment the vibration stops, so does the sound. - Striking a metal object — a taal or any sonorous metal rings when struck. - Blowing across an air column — in a bansuri the air inside the hollow pipe vibrates. - Striking a tuning fork on a rubber pad — the prongs vibrate; touch a prong to water and ripples spread out, proving the vibration (Activity 10.2). - Using the vocal cords — talking or singing vibrates the stretched flaps in the larynx; touch your throat while you speak and feel it. - Rubbing body parts — grasshoppers and crickets rub their wings or legs together. - Knocking, scratching or tapping — knock a desk (Activity 10.3), or tap two spoons together in air and again under water (Activity 10.4). - Rapid expansion of heated gas — a firecracker or a clap of thunder heats gas so fast that the sudden density disturbance is heard as a loud pulse. - Common thread: in every case the vibrating object is the source, and the sound ends when the vibration ends.

Question 3

Make a list of different types of musical instruments and identify their vibrating parts which produce sound. Threads of Curiosity What if your name were a tune instead of a word? In Kongthong, a village near Shillong in Meghalaya also known as the Whistling Village every person has a ‘tune name’ that can be sung or whistled. This unique tradition, called Jingrwai Iawbei, begins at birth when a mother composes a lullaby-like tune for her child. Let your friend gently knock or scratch on the desk. Listen carefully to the sound produced with your ear in the air.

Accepted Answer

- Every instrument sounds because some part of it vibrates — the instruments named in this chapter, with that part, are: - Bansuri (flute) — the air column inside the hollow pipe, set vibrating by blowing. - Tabla and mridangam — the stretched membrane of the drum head, whose vibration is shaped by the black syaahi patch at its centre. - Sitar, sarangi, veena, tanpura, ektara — the stretched strings; the sitar, sarangi and veena add extra strings to enrich the sound. - Taal — the metal discs themselves, which ring when struck because metals are sonorous. - Tuning fork — the two prongs (tines) of the U, struck against a rubber pad. - Rubber band stretched on an open box — the band, the simplest instrument you can make yourself. - Most instruments have more than one vibrating part — a plucked string also sets the body and the air inside it vibrating. - That is why a flute, an ektara and a tabla playing the same note at the same loudness still sound different: their timbre comes from the pattern of overtones their shape and material produce.

Question 4

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

Accepted Answer

(iii). - (iii) number of compressions per second. - Frequency is the number of density oscillations passing a point each second, so raising it means more compressions arrive every second. - (ii) is wrong: in a given medium the speed depends on the medium (and its temperature and humidity), not on the source or its frequency. - (i) is wrong: since v = ν × λ stays fixed, a larger ν forces a smaller λ. - (iv) is wrong: T = 1/(ν), so a higher frequency gives a shorter time period.

Question 5

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

Accepted Answer

5 Hz. - (ii) 5 Hz. - Each compression passing the point counts as one complete density oscillation there. - ν = ( number of oscillations )/( time taken ) = 20/(4 s ) = 5 Hz. - 5 Hz is below 20 Hz, so this would in fact be an infrasonic wave — inaudible to humans.

Question 6

Now, submerge the two metal spoons in water without touching the sides or bottom of the bucket and tap them against one another again (Fig. 10.6b). Do you again hear the sound produced? The sound of the submerged spoons tells you that the sound has reached you after travelling through water and air. If sound did not travel through liquids, would you have heard this sound? Sound can travel or propagate through solids, liquids and gases. The material through which sound propagates is called a medium. Sound propagates from its source to you through a medium. But suppose that there is no medium in the space between you and the source of the sound. A space where there is no medium (matter) is referred to as vacuum. Would you hear sound in vacuum? 10.2.1 Sound needs a medium to propagate A common experiment to show that sound needs a medium to propagate is the vacuum bell jar experiment shown in Fig. 10.7. An electric bell kept in a bell jar is switched on and the loudness of the sound is noted. As air is sucked out from the bell jar using a vacuum pump, the sound becomes fainter. Once a near vacuum is reached, almost no sound can be heard even though the bell can be seen ringing. When air is let back into the jar, the sound can be heard again and it gradually becomes as loud as before. This experiment shows that sound cannot propagate in vacuum. Sound needs a medium to propagate. The medium can be a solid, a liquid, or a gas. In outer space, there is a near vacuum, and thus, sound cannot propagate. Hence, astronauts in spacesuits doing spacewalks cannot directly hear each other speak or hear sounds like two metal objects clanking together. Instead, they communicate through special devices fitted into their spacesuits.

Accepted Answer

(ii). - (ii) Both A and R are true, and R is the correct explanation of A. - A is true: in the vacuum bell jar experiment the bell grows fainter as air is pumped out, and at near vacuum almost nothing is heard even though the bell is clearly seen ringing. - R is true: sound is a mechanical wave — it travels as compressions and rarefactions of the particles of a medium, so with no particles there is nothing to carry it. - R explains A: pumping out the air removes the medium, and removing the medium is exactly why the sound dies away. - The proof that this is the cause: let the air back in and the sound returns, gradually becoming as loud as before.

Question 7

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.

Accepted Answer

- Reverberation, not an echo. - An echo is heard as a separate sound only if the reflection arrives at least 0.1 s after the original. - Here the gap is 0.05 s, and 0.05 s < 0.1 s, so the brain cannot separate the two — no echo is heard. - Instead the quick multiple reflections from the walls blend with the original and make the sound persist after the source has stopped, which is reverberation. - This is why small rooms never echo: their walls are too close, and the reflections come back too soon.

Question 8

Give the slinky at your end a sharp push towards your friend and then quickly pull it back again (Fig. 10.8). Do you observe a disturbance created in the slinky which moves towards your friend?

Accepted Answer

(iii). - (iii) A is true, but R is false. - A is true: compressions and rarefactions — the regions of higher and lower density — really do travel forward through the medium, each particle passing the disturbance to the next by collision. - R is false: the particles themselves do not move forward with the wave; each only oscillates to and fro about its own mean position, parallel to the direction of propagation. - The slinky shows it directly: the marked turn stays where it is and merely oscillates, while the closely spaced and spread-out regions run the whole length of the slinky. - What travels forward is the disturbance and the energy it carries, not the matter of the medium.

Question 9

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? Density

Accepted Answer

(i) (a) (ii) (a). - (i) Wave (a) has the greater wavelength. (ii) Wave (a) has the smaller amplitude. - Both graphs plot density against distance on the same scales, so they can be compared square for square. - Wavelength is the distance between two consecutive crests: the crests in (a) are farther apart along the distance axis than those in (b), so λ a λ b. - Amplitude is the maximum change in density, i.e. the height of a crest above the average-density line: the curve in (a) rises less above that dashed line than the curve in (b), so (a) has the smaller amplitude. - So (a) is the long, gentle wave — lower frequency and less energy — and (b) is the short, tall one that carries more energy and sounds louder.

Question 10

Now, push and pull the slinky end multiple times in quick succession (The pulling and pushing of the end of the slinky is similar to the sound being produced continuously). Are a series of disturbances produced in the slinky? Do these disturbances move across the length of slinky? Does the mark on the slinky move back and forth parallel to the direction of the disturbance? Turns are closer together

Accepted Answer

(ii). - (ii) Energy carried by sound waves. - The air particles near the fork only oscillate about their mean positions and never travel to your ear, so (i) and (iv) are wrong — there is no stream of air moving across the room. - The fork's material stays in the fork, so (iii) is wrong. - Sound is a form of energy: the vibrating prongs hand energy to the neighbouring air particles, whose collisions pass it on from particle to particle until it sets your eardrum vibrating.

Question 11

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).

Accepted Answer

- Compressions (C) are the crowded regions and rarefactions (R) the thinly spread ones — mark them alternately, C R C R …, along both Fig. 10.17 (a) and (b). - In (a) and (b): write C wherever the dots representing the particles are packed closer than average, and R wherever they are more spread out than average. - For graphs (c) and (d): label the x-axis "Distance" and the y-axis "Density". - Draw a horizontal dashed line across each graph for the average density — this is the level the curve oscillates about. - On each graph draw a smooth wavy curve with a crest (maximum density) directly above every C and a trough (minimum density) directly above every R, the curve cutting the dashed line midway between them. - Keep both graphs on the same distance scale: the strip whose C's are closer together must give the curve with crests closer together, i.e. the shorter wavelength, and the other strip the longer-wavelength curve. - The height of a crest above the dashed line (and the depth of a trough below it) is the density amplitude — draw it the same on both unless one strip is visibly more crowded than the other.

Question 12

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.

Accepted Answer

A - Green curve; B - Red curve; C - Blue curve. - A is the green curve, B is the red curve and C is the blue curve. - All three waves travel in the same medium, so they share one speed v, and v = ν × λ gives λ = v/(ν) — the highest frequency must have the shortest wavelength. - A has the maximum frequency, so mark A on the curve whose crests are packed closest together — the green one. - C has the minimum frequency, so mark C on the curve with the widest crest-to-crest spacing — the blue one. - B is left with the in-between spacing, the red curve, giving λ A < λ B < λ C while ν A ν B ν C. - Write the letters A, B and C directly on their three curves in Fig. 10.31.

Question 13

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

Accepted Answer

- Draw a density-versus-distance graph whose crests stand 3 units above the average-density line and which repeats every 4 cm. It must contain: - An x-axis labelled "Distance (cm)", marked 0, 1, 2, 3, 4, 5, … up to at least 8 cm. - A y-axis labelled "Density (units)", marked from -3 to +3 about the average value. - A horizontal dashed line right across the graph for the average density — the level the curve oscillates about. - A smooth wave whose crests sit exactly 3 units above the dashed line and whose troughs dip 3 units below it; that 3 units is the density amplitude. - A crest-to-crest spacing of λ = 4 cm: put crests at 0, 4 and 8 cm, troughs midway between them at 2 and 6 cm, and the curve crossing the dashed line at 1, 3, 5 and 7 cm. - C marked under each crest and R under each trough, with a double-headed arrow between two consecutive crests labelled λ = 4 cm. - At least two full cycles, so that the repetition is visible.

Question 14

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?

Accepted Answer

- Yes — the thin rubber band vibrates faster than the thick one. - Faster vibration means more density oscillations reach your ear each second, so the thin band gives the higher frequency and the shriller, higher-pitched sound. - Its time period is shorter, because frequency and time period are inversely related: ν = 1/T, so a larger ν forces a smaller T. - The thick band is the opposite — lower frequency, longer time period, deeper sound. - The chapter gives the same rule in the human voice: during adolescence boys' vocal cords lengthen and thicken, vibrate less frequently, and the voice 'deepens'. - Tightening either band also raises its frequency, which is the change you hear in step 6 of Activity 10.1.

Question 15

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?

Accepted Answer

- Two errors: there should be no sound at all, and even where sound is possible it could never arrive together with the flash. - Error 1 — sound in space. Space is a near vacuum, and sound is a mechanical wave that needs a material medium to carry its compressions and rarefactions. With no particles, the explosion would be completely silent to a distant observer. - This is the same reason astronauts on a spacewalk cannot hear each other speak or hear metal clanking, and must use radio devices in their spacesuits. - Error 2 — the timing. Light travels at about 300000 km s^-1 while sound in air manages only about 340 m s^-1, so sound always lags far behind the flash — as thunder does behind lightning. Showing them at the same instant is wrong even on Earth. - The flash itself is correct: light is not a mechanical wave and travels through vacuum, which is how starlight reaches us.

Question 16

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?

Accepted Answer

1200 oscillations. - 1200 oscillations per minute. - Frequency is the number of oscillations per unit time, so ν = 20 Hz means the piston completes 20 oscillations every second. - One minute is 60 s, so oscillations = 20 s^-1 × 60 s = 1200. - Each of those 1200 oscillations sends out one compression and one rarefaction into the tube.

Question 17

For the sound wave represented by the graph shown in Fig. 10.19, what is half of its wavelength? The amplitude The amount of sound energy passing through a unit area perpendicular

Accepted Answer

1.5 cm. - Half the wavelength is 1.5 cm, so the full wavelength is 3.0 cm. - On the distance axis of Fig. 10.19 the marks 0, 1.5, 3.0, 4.5 cm fall alternately on a crest and a trough. - Wavelength is the distance between two consecutive crests (or two consecutive troughs): λ = 3.0 cm - 0 = 3.0 cm. - (λ)/2 = (3.0 cm )/2 = 1.5 cm — which is exactly the crest-to-next-trough spacing you can read straight off the graph.

Question 18

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.

Accepted Answer

0.01 s. - Time period T = 0.01 s. - From v = ν × λ, the frequency is ν = v/(λ) = (344 m s^-1)/(3.44 m ) = 100 Hz. - From ν = 1/T, the time period is T = 1/(ν) = 1/(100 Hz ) = 0.01 s.

Question 19

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?

Accepted Answer

3812.5 m 65. - The wreck lies about 3812.5 m down. - The 5 s is the round trip, so the signal takes (5 s )/2 = 2.5 s to reach the wreck. - depth = v × t = 1525 m s^-1 × 2.5 s = 3812.5 m. - In one step: d = (v t)/2 = (1525 m s^-1 × 5 s )/2 = 3812.5 m.

Question 20

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.

Accepted Answer

(i) 75:17 (ii) 10:3. - (i) water: air = 75: 17. (ii) steel: water = 10: 3. - From Table 10.1 at 15 ºC: steel 5000 m s^-1, water 1500 m s^-1, air 340 m s^-1. - (i) v water v air = 1500/340 = 150/34 = 75/17 ≈ 4.4 — sound is about 4–5 times faster in water than in air. - (ii) v steel v water = 5000/1500 = 10/3 ≈ 3.3 — and steel is about 15 times faster than air. - Both ratios confirm the chapter's ordering: sound is fastest in solids, slower in liquids and slowest in gases.

Question 21

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.

Accepted Answer

0.007 s. - About 0.007 s. - The pulse goes to the obstacle and comes back, so the path length is 2 × 1.2 m = 2.4 m. - t = ( distance )/( speed ) = (2.4 m )/(345 m s^-1) = 0.00696 s ≈ 0.007 s. - The 40 kHz is not needed for the time — it only tells you the wave is ultrasonic (above 20 kHz), so the driver never hears the pulse itself, only the warning beep the system produces.

Question 22

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.) 10.6.4 Human perception of sound The physical properties of sound, such as time period, wavelength, frequency, amplitude and speed are well-defined and can be measured. However, how we experience sound is subjective. Human perception of sound is described by terms, such as loudness and pitch which do not have simple relations with the physical properties. Pitch How frequency is perceived by humans is called pitch. Sounds perceived to be shrill, such as a whistle or a siren are said to have high pitch, while deep sounds like thunder or an aircraft rumble have low pitch. In general, high pitch sounds have higher frequency and low pitch sounds have lower frequency, although the exact mathematical relation is complicated. Threads of Curiosity Everyone’s voice sounds unique and you can often instantly recognise your teacher or friends calling out your name. Why do our voices sound different? Male, female, and children’s voices differ not just in their frequency but also on how the sound is shaped by the throat, mouth, and nasal cavities. During adolescence, boys’ vocal cords of boys lengthen and thicken, vibrating less frequently thus, ‘deepening’ their voice. Let us carry out an activity to listen and appreciate sounds at different frequencies. Activity 10.8: Let us experiment (demonstration activity) This activity is recommended to be performed as a classroom group activity facilitated by teacher. 1. Open a mobile app that can generate sounds. 2. Set the frequency to 100 Hz, tap ‘play’, and listen carefully. 3. Increase the frequency in steps of 100 Hz up to 1000 Hz and describe how the sound changes. 4. Next, set the frequency to 50 Hz. Reduce the frequency till about 20 Hz or the point where you cannot hear the sound anymore.

Accepted Answer

0.932 s; Yes. - Time difference = 0.932 s, and yes — Gunjan could hear the two as separate sounds. - Through air: t air = d v air = (340 m )/(340 m s^-1) = 1 s. - Through steel: t steel = (340 m )/(5000 m s^-1) = 0.068 s. - Δ t = 1 s - 0.068 s = 0.932 s. - 0.932 s is far more than the 0.1 s the ear needs, so the two arrivals are heard separately. - The knock reaches her through the steel first (sound is fastest in solids), and the same knock arrives through the air nearly a second later.

Question 23

The speed of sound in air is about 331 m s^-1 at 0 ºC and nearly 344 m s^-1 at 22 ºC. 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 ºC to 0 ºC? Assume that all other conditions remain unchanged.

Accepted Answer

- The sound of the thunder takes about 0.2 s extra to cover the same 1720 m once the air cools from 22^° C to 0^° C. - Section 10.6.3 gives the tool: speed = distance ÷ time, used in Example 10.3 as distance = v × t; rearranged, time = ( distance )/( speed ). - At 22^° C the speed is 344 m s^-1 (Section 10.6.3, repeated in the question), so t 22 = (1720 m )/(344 m s^-1) = 5.00 s. - At 0^° C the speed is 331 m s^-1, so t 0 = (1720 m )/(331 m s^-1) = 5.196 s ≈ 5.2 s. - The extra time is the difference: Δ t = t 0 - t 22 = 5.196 s - 5.00 s = 0.196 s ≈ 0.2 s. - Direction check against the chapter: Section 10.6.3 states that as the temperature is increased the speed of sound increases, so cooling the air lowers the speed, and a slower wave needs longer for the same distance. The difference therefore has to come out positive and small compared with the 5 s journey itself - which 0.196 s is. - The book's printed Answer Key gives 331 s, which does not agree with Section 10.6.3 (nor with the question's own stem): 331 there is the speed of sound in dry air at 0^° C, in m s^-1, not a time. The key entry appears to have copied that speed and re-labelled it as seconds. - Reasoning beyond the chapter, to show the printed figure cannot be rescued by any reading of the question: the entire trip at 0^° C lasts only 5.2 s, so no delay inside this problem can be 331 s; and 331 s of travel at roughly 340 m s^-1 would be 331 s × 340 m s^-1 = 112540 m ≈ 113 km, not the 1720 m the question sets. The chapter's own two speeds settle it the other way, at 0.196 s ≈ 0.2 s.

Question 24

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. Modern auditoriums and large concert halls are architecturally

Accepted Answer

34.3 m. - The reflecting surface must be at least 34.3 m away. - The sound must travel to the surface and back, so in 0.2 s it covers twice the distance d. - 2d = v × t = 343 m s^-1 × 0.2 s = 68.6 m. - d = (68.6 m )/2 = 34.3 m. - Any nearer and the echo returns sooner than 0.2 s, which the experiment does not allow.

Question 25

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.

Accepted Answer

0.04 m; 8500 Hz. - Wavelength λ = 0.04 m (4 cm) and frequency ν = 8500 Hz. - The 8 cm marked in Fig. 10.32 spans two complete cycles — two compressions and two rarefactions — so one wavelength is λ = (8 cm )/2 = 4 cm = 0.04 m. - From v = ν × λ, ν = v/(λ) = (340 m s^-1)/(0.04 m ) = 8500 Hz. - 8500 Hz lies inside the audible range of 20 Hz to 20 kHz, so this sound can be heard.

Question 26

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 Quest Continues … Sound helps us explore places and phenomenon beyond human hearing. Space probes have recorded the first sounds from Mars, scientists are timing the sound of distant earthquakes to measure tiny changes in ocean temperature to understand the Earth’s changing climate, biologists are using the buzz of mosquitoes to identify disease-carrying mosquitoes, and researchers are listening to the tiny crackles produced by microbes in the soil to study soil health and biodiversity. As technology improves, sound is becoming an even more powerful tool to explore planets, living organisms, and the hidden activities of nature.

Accepted Answer

3000 m. - The ocean is 3000 m deep at that location. - The 4 s is the time for the signal to go down and come back, so the one-way time is (4 s )/2 = 2 s. - depth = v × t = 1500 m s^-1 × 2 s = 3000 m. - This is sonar — ultrasonic pulses are used because sound travels far in water, while light does not.

Question 27

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. Density

Accepted Answer

0.025 m, 0.05 m; 13800 Hz, 6900 Hz. - Wave A: λ A = 0.025 m, ν A = 13800 Hz. Wave B: λ B = 0.05 m, ν B = 6900 Hz. - Read each wavelength crest-to-consecutive-crest off the shared distance axis of Fig. 10.33: A repeats every 2.5 cm = 0.025 m, B every 5.0 cm = 0.05 m. - ν A = v/(λ A) = (345 m s^-1)/(0.025 m ) = 13800 Hz. - ν B = v/(λ B) = (345 m s^-1)/(0.05 m ) = 6900 Hz. - B's wavelength is exactly twice A's, so its frequency is exactly half — the two share one speed, as v = ν λ requires.

Question 28

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?

Accepted Answer

2:9. - v air: v water = 2: 9. - Both sources sit the same horizontal distance d from the cliff face, and each sound travels there and back, covering 2d. - Times to return: t A = 2d v air and t B = 2d v water. - Given t A = 4.5 t B: 2d v air = 4.5 × 2d v water, and 2d cancels from both sides. - 1 v air = 4.5 v water ⇒ v air v water = 1/4.5 = 2/9. - So sound is 4.5 times faster in water than in air — matching the chapter's statement that sound travels about 4–5 times faster in water.

Revise, Reflect, Refine 10.1 (part 1 of 21)

Exercise 10.1