Question

A source emits sound with a frequency of 1,000 Hz. It is moving at 20 m/s toward a stationary reflecting wall. If the speed of sound is 340 m/s, an observer at rest directly behind the source hears a beat frequency of

Answer 1

Answer:

the beat frequency heard by the observed is 58.8 Hz.

Explanation:

Given;

frequency of the source, Fs = 1,000 Hz

velocity of the source, Vs = 20 m/s

velocity of sound, V = 340 m/s

The frequency heard by the observer behind the source is calculated by applying Doppler effect;

$F_s = F_o [\frac{v}{v-v_s} ]\\\\1000 = F_o [\frac{340}{340 -20} ]\\\\1000 = 1.0625F_o\\\\F_o = 941.2 \ Hz$

The beat frequency heard by the observed = 1000 - 941.2 = 58.8 Hz.

Therefore, the beat frequency heard by the observed is 58.8 Hz.

Answer 2

Answer:

• An observer at rest directly behind the source hears a beat frequency of $117.65Hz$

Explanation:

$N_1$ = frequency observed by observer when source goes away from the observer

$N_2$ = when source coming towards observer

$N_1 = m(\frac{V-V_o}{V})\\\\N_1 = 1000(\frac{340-20}{340})\\\\N_1 = 941.17$

$N_2 = m(\frac{V+V_o}{V})\\\\N_2 = 1000(\frac{340+20}{340})\\\\N_2 = 1058.82$

Therefore,

$N_2 - N_1 = 1058.82 - 941.17$

Beat frequency = $117.65Hz$

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