Two cars A and B are moving away from each other in opposite directions. Both the cars are moving with a speed of 20 ms^-1 with respect to the ground. If an observer in car A detects a frequency 2000 Hz of the sound coming from car B, what is the natural frequency of the sound source in car B?

(Speed of sound in air = 340 ms^-1)

Concept:

Doppler Effect:

The person hearing the sounds, are called the observer or listener and the thing emitting the sound is called the source.

The Doppler Effect is the change in the observed frequency of a wave when the source or the detector moves relative to the transmitting medium.

Here, v₀ and v_s is taken when the cars are approaching each other.

According to Doppler Effect, when both source and observer moves away from each other

\(f' = {f_0}\frac{{v - {v_0}}}{{v + {v_s}}}\)

Calculation:

Given,

Speed of car A (Observer velocity), v₀ = 20 ms^-1

Speed of car B (Source velocity), v_s = 20 ms^-1

Apparent Frequency (frequency observed in car A, f' = 2000Hz

Velocity of the sound in air, v = 340 ms^-1

According to Doppler Effect,

\(f' = {f_0}\frac{{v - {v_0}}}{{v + {v_s}}}\)

\(2000 = {f_0}\frac{{340 - 20}}{{340 + 20}}\)

\(2000 = {f_0}\left( {\frac{{320}}{{360}}} \right)\)

\(2000 = {f_0}\left( {\frac{8}{9}} \right)\)

\({f_0} = \frac{{2000 \times 9}}{8}\)

∴ f₀ = 2250 Hz

Thus, the natural frequency of the sound source in car B is 2250 Hz.