Propagation of Sound Waves Class 9 ICSE Important Questions with Answers

Answer: (b) longitudinal mechanical waves.

Sound travels as compressions and rarefactions along the direction of propagation and needs a material medium, so it is a longitudinal mechanical wave.

Answer: (d) vacuum.

Sound requires a material medium (solid, liquid or gas) to propagate because it is carried by the vibration of particles; a vacuum has no particles.

Answer: (a) $330\ \text{m s}^{-1}$330 m s^-1.

$v=f\lambda=500\times0.66=330\ \text{m s}^{-1}$v=fλ=500×0.66=330 m s^-1.

Answer: (c) supersonic.

Supersonic describes a speed greater than the speed of sound. Ultrasonic describes a frequency of sound above $20\,000\ \text{Hz}$20 000 Hz, too high for humans to hear. The two words are often confused.

Answer: (b) energy. The particles of the medium only vibrate about their mean positions; they pass the disturbance, and the energy, on to the next particles.

Answer: (a) Both A and R are true and R is the correct explanation of A. Lightning and thunder are produced together, but light (about $3\times10^8\ \text{m s}^{-1}$3×10⁸ m s^-1) reaches us almost instantly while sound (about $340\ \text{m s}^{-1}$340 m s^-1) takes a noticeable time.

(i) Infrasonic: below $20\ \text{Hz}$20 Hz; produced by earthquakes and heard by elephants and whales.

(ii) Audible (sonic): $20\ \text{Hz}$20 Hz to $20\,000\ \text{Hz}$20 000 Hz, the range of human hearing.

(iii) Ultrasonic: above $20\,000\ \text{Hz}$20 000 Hz; produced and heard by bats and dolphins, and used in ultrasound scanners.

It proves that sound requires a material medium to travel and cannot pass through a vacuum.

As the air is removed, there are fewer and fewer particles to carry the compressions and rarefactions of the sound wave, so the sound grows fainter. When the jar is nearly evacuated almost no sound is heard, even though the bell hammer is still seen striking.

Infrasonic describes the frequency of a sound: below $20\text{ Hz}$20 Hz, too low for humans to hear (for example, sounds from earthquakes).

Subsonic describes a speed: less than the speed of sound in that medium (for example, a passenger aircraft cruising at about $900\text{ km h}^{-1}$900 km h^-1, below the roughly $1200\text{ km h}^{-1}$1200 km h^-1 speed of sound in air).

Distance $=$= speed of sound $\times$× time $=340\times4=1360\ \text{m}$=340×4=1360 m, i.e. about $1.36\ \text{km}$1.36 km.

Light travels at about $3\times10^8\ \text{m s}^{-1}$3×10⁸ m s^-1, so it covers $1360\ \text{m}$1360 m in about $4.5$4.5 millionths of a second. This is negligible compared with $4\ \text{s}$4 s, so we take the flash as seen at the instant it happens.

The speed of sound depends on the elasticity and the closeness of the particles of the medium.

In solids the particles are very close together and tightly bound, so a vibration is passed on very quickly from one particle to the next; solids are also highly elastic. Hence sound travels fastest in solids.

In gases the particles are far apart and loosely held, so the disturbance is transferred slowly and the elasticity is low. Hence sound travels slowest in gases. Liquids lie in between: $v_{solid}>v_{liquid}>v_{gas}$v_solid>v_liquid>v_gas.

(a) Using $v=f\lambda$v=fλ,
$\lambda=\frac{v}{f}=\frac{340}{170}=2\ \text{m}.$λ=v/f=340/170=2 m.
The wavelength is $2\ \text{m}$2 m.

(b) At $f=340\ \text{Hz}$f=340 Hz (same speed, as the medium is unchanged):
$\lambda=\frac{340}{340}=1\ \text{m}.$λ=340/340=1 m.
The wavelength is halved to $1\ \text{m}$1 m, since wavelength is inversely proportional to frequency at constant speed.

Wavelength is the distance between two consecutive crests (or troughs) of a transverse wave, and between two consecutive compressions (or rarefactions) of a longitudinal wave.

(a) In a longitudinal sound wave the particles vibrate back and forth along the direction of travel, creating regions of high pressure (compressions, C) and low pressure (rarefactions, R).

(b) Wavelength is the distance between two consecutive compressions (or two consecutive rarefactions); equivalently, the distance covered by the wave in one complete vibration.

(c) $v=f\lambda=256\times1.32=337.9\ \text{m s}^{-1}\approx338\ \text{m s}^{-1}$v=fλ=256×1.32=337.9 m s^-1≈338 m s^-1.

The speed of sound in air is about $338\ \text{m s}^{-1}$338 m s^-1.

(a) (i) Temperature: speed increases as temperature rises (by about $0.6\ \text{m s}^{-1}$0.6 m s^-1 for each $1^\circ\text{C}$1° C), because the air particles move faster.
(ii) Humidity: speed increases in moist air, because moist air is less dense than dry air.
(iii) Pressure: at constant temperature, a change of pressure has no effect on the speed of sound in a gas.

(b) Any two: ultrasound scanning of the body (e.g. of an unborn baby); SONAR, to find the depth of the sea or locate submarines and shoals of fish; cleaning delicate parts; detecting cracks in metal blocks.

(c) Sound travels much faster in solids (about $5000\ \text{m s}^{-1}$5000 m s^-1 in steel) than in air (about $340\ \text{m s}^{-1}$340 m s^-1), so the sound through the rails arrives first.

(i) Sonar depends on the reflection of sound (echo) from the sea bed.

(ii) Sound travels efficiently over long distances through water with little absorption, whereas light is strongly absorbed and scattered by water and cannot penetrate to great depths. Hence sound is used.

(iii) The sound travels down and back, so it covers $2d$2d in $4\ \text{s}$4 s:
$2d=v\times t=1500\times4=6000\ \text{m}\ \Rightarrow\ d=\frac{6000}{2}=3000\ \text{m}.$2d=v× t=1500×4=6000 m ⇒ d=6000/2=3000 m.
The sea bed is $3000\ \text{m}$3000 m (3 km) deep.

(iv) A greater depth means the sound has to travel a longer total path, so the return time would increase (time is directly proportional to depth for the same speed).