10th Science Unit 5 Acoustics Book Back Answers

Sound waves are longitudinal waves.

In a longitudinal wave, the particles of the medium vibrate along the direction of propagation of the wave, producing compressions and rarefactions.

The velocity of sound in a gas depends on the density and the temperature of the gas, not on the pressure (pressure and density change together at constant temperature).

Since the temperature is unchanged, the velocity remains $330\,\mathrm{ms^{-1}}$.

Audible waves have frequencies between 20 Hz and 20,000 Hz (20 kHz).

Among the options only 20 kHz lies within this audible range.

Velocity of sound in a gas is directly proportional to the square root of its temperature: $v \propto \sqrt{T}$; pressure has no effect.

When $T$ is doubled, $v' = v\sqrt{2} = 330 \times \sqrt{2}\,\mathrm{ms^{-1}}$.

Use $V = n\lambda$, so $\lambda = \dfrac{V}{n}$.

$\lambda = \dfrac{344}{1.25 \times 10^{4}} = 0.02752\ \mathrm{m}$.

When sound is reflected back into the same medium, the medium is unchanged, so its speed stays the same.

Frequency is decided by the source, so it does not change; since $V = n\lambda$ is unchanged, the wavelength also stays the same.

Minimum distance to hear an echo is $d = \dfrac{v \times 0.1}{2} = \dfrac{v}{20}$.

$d = \dfrac{500}{20} = 25\ \mathrm{m}$.

Sound is produced by vibrations — the rapid to and fro motion of a particle about its mean (rest) position.

In a longitudinal wave the particles of the medium vibrate along the direction of propagation of the wave.

If the energy travels from south to north, the particles vibrate along the same south–north line.

Listener at rest, source moving towards the listener: $n' = \dfrac{v}{v - v_s}\,n$

Substituting: $n' = \dfrac{330}{330 - 33} \times 450 = \dfrac{330}{297} \times 450$

$n' = 500\ \mathrm{Hz}$ (apparent frequency is more than the actual frequency).

Observer at rest, source approaching: $n' = \dfrac{v}{v - v_s}\,n$

Here $v = 1220$ km/h, $v_s = 40$ km/h, $n = 2000$ Hz (the units of $v$ and $v_s$ are the same, so no conversion is needed).

$n' = \dfrac{1220}{1220 - 40} \times 2000 = \dfrac{1220}{1180} \times 2000$

$n' \approx 2067.8\ \mathrm{Hz}$

Sound is a longitudinal wave and needs a material medium (solid, liquid or gas) for its propagation.

On the Moon there is no air, so no sound can be heard — proving sound cannot travel through vacuum.

Infrasonic waves are sound waves of frequency below 20 Hz, which the human ear cannot hear.

Waves produced during an earthquake are given in the text as an example of infrasonic waves.

The velocity of sound in a gas is directly proportional to the square root of its temperature, $v \propto \sqrt{T}$.

It is given by $v_T = (v_0 + 0.61\,T)$ m s$^{-1}$, so velocity changes by 0.61 m s$^{-1}$ for every 1 °C rise.

Velocity of sound depends on the elasticity of the medium; solids are most elastic and gases are least elastic.

Hence $v_S > v_L > v_G$; e.g. sound travels at 331 m s$^{-1}$ in air at 0 °C but 1493 m s$^{-1}$ in water.

Infrasonic waves have frequency below 20 Hz, so 10 Hz is infrasonic → (c)

The principle of echo is used in obstetric ultrasonography to image the fetus → (d)

Ultrasonic waves have frequency greater than 20 kHz, so 22 kHz is ultrasonic → (b)

In a longitudinal sound wave the high pressure regions are called compressions → (a)

Assertion check: At constant temperature, if the pressure of a gas is increased, its density increases in exactly the same proportion, so the ratio P/d is unchanged. Since v ∝ √(P/d), the speed of sound in a gas is NOT altered by a change of pressure at constant temperature. (This is why in MCQ 2, making the pressure 4 times leaves the velocity at 330 ms⁻¹.) Hence the assertion is FALSE.

The factors that really decide the speed of sound in a gas are its density, temperature and relative humidity (v ∝ 1/√d, v ∝ √T, v increases with humidity).

Since the assertion is false, the only choice available that begins with 'Assertion is false' is option (d). Hence the answer keyed for this question is (d).

Caution while writing the answer: the reason as printed ('speed of sound in a gas is proportional to the square of the pressure') is itself not an accurate statement — the correct relation is v ∝ √(P/d), i.e. a square-root dependence. So the reason is at best only partly acceptable; the choice is made on the strength of the assertion being false.

Assertion check: Velocity of sound is maximum in solids because they are more elastic than liquids and gases; gases are least elastic, so $v_S > v_L > v_G$. The assertion is true.

Reason check: It is a true fact that solids have a much greater density than gases.

But density is not the reason for the higher speed — the speed of sound is inversely proportional to the square root of density, so a greater density by itself would lower the speed. The real cause is the much larger elastic modulus of solids.

Hence both statements are true, but the reason is not the correct explanation of the assertion.

A longitudinal wave is a wave in which the particles of the medium vibrate along (parallel to) the direction of propagation of the wave.

This back and forth vibration produces a series of high pressure regions called compressions and low pressure regions called rarefactions.

Sound waves are longitudinal waves and they can travel through solids, liquids and gases.

Sound waves that the human ear can hear are called audible waves.

Their frequency ranges between 20 Hz and 20,000 Hz (20 kHz).

They are produced by vibrating bodies such as vocal cords and stretched strings.

The persistence of hearing for the human ear is 0.1 s, so the echo must reach the ear at least 0.1 s after the original sound.

Sound travels to the reflector and back, a distance $2d$, so $d = \dfrac{v t}{2} = \dfrac{v \times 0.1}{2} = \dfrac{v}{20}$.

Taking the velocity of sound in air as $344\,\mathrm{ms^{-1}}$, $d = \dfrac{344}{20} = 17.2\,\mathrm{m}$.

Given: wavelength $\lambda = 0.20\,\mathrm{m}$, velocity $v = 331\,\mathrm{ms^{-1}}$.

Formula: $v = n\lambda$, so $n = \dfrac{v}{\lambda}$.

Substituting: $n = \dfrac{331}{0.20}$.

$n = 1655\,\mathrm{Hz}$.

Ultrasonic waves have frequency greater than 20 kHz and cannot be heard by the human ear.

But certain creatures can detect them.

Examples: bats, dogs and dolphins (mosquitoes also detect them).

The velocity of sound in a gas depends on the relative humidity of the air.

On a rainy day the amount of water vapour in air is high, so the relative humidity is high.

When humidity increases, the speed of sound increases; hence sound travels faster and can be heard clearly from long distances during rainy seasons.

On a dry day the humidity is low, so sound travels comparatively slower.

Sound is produced by vibrations, and an empty vessel has a large volume of air column inside it.

When struck, the sound waves inside the empty vessel undergo multiple reflections from its walls and the air column vibrates freely.

These repeated reflections add up, so a louder sound (with echo-like reverberation) is heard.

In a filled vessel the liquid or solid damps the vibrations and prevents free reflection of sound inside, so less sound is produced.

Given: $v_0 = 331\,\mathrm{ms^{-1}}$, temperature $T = 46^\circ C$.

Formula: $v_T = (v_0 + 0.61\,T)\ \mathrm{ms^{-1}}$, since velocity changes by $0.61\,\mathrm{ms^{-1}}$ for each degree celsius rise.

Substituting: $v_T = 331 + (0.61 \times 46)$.

$v_T = 331 + 28.06 = 359.06\,\mathrm{ms^{-1}}$.

Sound reflected from a plane surface keeps the same intensity, but reflection from a curved surface changes the intensity of the reflected waves.

A curved (concave / parabolic) ceiling converges the reflected sound waves and focuses them towards the audience.

So the sound reaches every part of the hall with sufficient intensity and no portion of the hall is left silent.

Hence the quality and loudness of sound heard by the audience is improved, which is why concert hall ceilings are made curved.

Doppler effect occurs only when the distance between source and listener changes due to relative motion.

Case 1: When both the source (S) and the listener (L) are at rest — there is no relative motion, so no Doppler effect.

Case 2: When S and L move in such a way that the distance between them remains constant — the apparent frequency equals the source frequency.

(Also, when S and L move in mutually perpendicular directions, or when the source is at the centre of the circle along which the listener moves.)

Given: frequency $n = 200\,\mathrm{Hz}$, velocity $v = 400\,\mathrm{ms^{-1}}$.

Formula: $v = n\lambda$, so $\lambda = \dfrac{v}{n}$.

Substituting: $\lambda = \dfrac{400}{200}$.

Result: $\lambda = 2\,\mathrm{m}$.

Given: time taken by sound $t = 9.8\,\mathrm{s}$, speed of sound $v = 330\,\mathrm{ms^{-1}}$ (light from the flash reaches instantly).

Formula: distance $d = v \times t$.

Substituting: $d = 330 \times 9.8$.

Result: $d = 3234\,\mathrm{m}$, which is the height of the cloud.

Given: frequency of sound $n = 600\,\mathrm{Hz}$ (the distance 400 m is not needed).

The time between two successive compressions is one time period $T$.

Formula: $T = \dfrac{1}{n}$.

Substituting: $T = \dfrac{1}{600} = 1.67 \times 10^{-3}\,\mathrm{s}$.

Given: total time for the wave to go down and come back $t = 1.6\,\mathrm{s}$, velocity in sea water $v = 1400\,\mathrm{ms^{-1}}$.

The wave travels a total distance $2d$ (ship to sea bottom and back).

Formula: $v = \dfrac{2d}{t}$, so $d = \dfrac{v \times t}{2}$.

Substituting: $d = \dfrac{1400 \times 1.6}{2} = \dfrac{2240}{2}$.

Result: $d = 1120\,\mathrm{m}$.

Given: distance between the walls $d_1 + d_2 = 680\,\mathrm{m}$, echo times $t_1 = 0.9\,\mathrm{s}$ and $t_2 = 1.1\,\mathrm{s}$.

For the first wall: $2d_1 = v \times 0.9$; for the second wall: $2d_2 = v \times 1.1$.

Adding: $2(d_1 + d_2) = v(0.9 + 1.1) = v \times 2$.

Substituting: $2 \times 680 = v \times 2$, so $1360 = 2v$.

Result: $v = 680\,\mathrm{ms^{-1}}$.

Given: distance between the boats $d = 4.5\,\mathrm{km} = 4500\,\mathrm{m}$, time $t = 3\,\mathrm{s}$.

The signal travels straight from one boat to the other, so it covers $d$ (not $2d$).

Formula: $v = \dfrac{d}{t}$.

Substituting: $v = \dfrac{4500}{3}$.

Result: $v = 1500\,\mathrm{ms^{-1}}$.

Given: total time for the signal to return $t = 1\,\mathrm{s}$, speed of sound in water $v = 1450\,\mathrm{ms^{-1}}$.

The signal covers a total distance $2d$ (ship to bottom and back).

Formula: $d = \dfrac{v \times t}{2}$.

Substituting: $d = \dfrac{1450 \times 1}{2}$.

Result: $d = 725\,\mathrm{m}$.

Wavelength is given by $\lambda = v/n$, so for the same frequency $n$, the wave with greater speed has the longer wavelength.

Sound travels in air at about $340\ \mathrm{ms^{-1}}$, while light travels at $3\times10^{8}\ \mathrm{ms^{-1}}$ — nearly a million times faster.

Hence for equal frequency, the light wave has the much longer wavelength.

To hear an echo, the reflected sound must reach the ear at least 0.1 s after the original sound, so the distance must be at least $d = v \times 0.1/2 = v/20$.

On a hotter day the velocity of sound increases, since $v \propto \sqrt{T}$ (velocity rises by about $0.61\ \mathrm{ms^{-1}}$ for each $1^\circ$C rise).

With the distance unchanged, the greater velocity means the sound returns in a shorter time $t = 2d/v$.

If this time falls below 0.1 s, the echo merges with the original sound (persistence of hearing) and no echo is heard. So on a hotter day the echo becomes fainter or is not heard at all, unless the distance is well above the minimum $v/20$.