Propagation of Sound Waves — ICSE Class 9 Physics Important Questions

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=f=500×0.66=330 m s^-1.

Answer: (c) quality (timbre).

Same loudness means same amplitude and same pitch means same frequency, so the difference must be in quality, which is determined by the waveform (the overtones present).

Answer: (a) The reflected sound must reach the ear at least $0.1\ \text{s}$0.1 s after the original for it to be heard separately as an echo. This requires a minimum distance of about $17\ \text{m}$17 m, so R correctly explains A.

Loudness: the property by which a loud sound is distinguished from a faint one; it depends on the amplitude of the wave (greater amplitude means louder sound).

Pitch: the property by which a shrill sound is distinguished from a flat one; it depends on the frequency of the wave (higher frequency means higher pitch). Thus loudness relates to amplitude while pitch relates to frequency.

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.

Given: time for echo $t=3\ \text{s}$t=3 s, speed of sound $v=340\ \text{m s}^{-1}$v=340 m s^-1.

In time $t$t the sound travels to the cliff and back, covering a total distance $2d$2d:
$2d=v\times t=340\times3=1020\ \text{m}.$2d=v× t=340×3=1020 m.

Therefore
$d=\frac{1020}{2}=510\ \text{m}.$d=1020/2=510 m.

The cliff is $510\ \text{m}$510 m from the person.

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=f,
$\lambda=\frac{v}{f}=\frac{340}{170}=2\ \text{m}.$=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):
$\lambda=\frac{340}{340}=1\ \text{m}.$=340/340=1 m.
The wavelength is halved to $1\ \text{m}$1 m, since wavelength is inversely proportional to frequency at constant speed.

(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=f=256×1.32=337.9 m s^-1338 m s^-1.

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

(a) An echo is the sound heard again after it is reflected from a distant hard surface such as a cliff or wall. To hear a distinct echo the reflected sound must reach the ear at least $0.1\ \text{s}$0.1 s after the direct sound, because the sensation of sound persists in the ear for about $0.1\ \text{s}$0.1 s.

(b) In $0.1\ \text{s}$0.1 s the sound travels to the surface and back, a total distance $2d$2d:
$2d=v\times t=340\times0.1=34\ \text{m}\ \Rightarrow\ d=17\ \text{m}.$2d=v× t=340×0.1=34 m d=17 m.
The minimum distance is $17\ \text{m}$17 m.

(c) $2d=v\times t=340\times0.5=170\ \text{m}$2d=v× t=340×0.5=170 m, so
$d=\frac{170}{2}=85\ \text{m}.$d=170/2=85 m.
The wall is $85\ \text{m}$85 m away.

(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).