Question

The frequency changes by 10% as the source approaches a stationary observer with constant speed Vs. What should be the percentage change in frequency as the source recedes from the observer with the same speed? Given that Vs ≪ v(v is the speed of sound in air).

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

When the source approaches the observer,
$f_1=f\left(\frac{v}{v-v_s}\right)=f{(1-\frac{v_s}{v})}^{-1}\approx f(1+\frac{v_s}{v})or\left(\frac{f_1-f}{f}\right)\times 100=\frac{v_s}{v}\times 100=\mathrm{10.........}\left(1\right)$
In the second case, when the source recedes from the observer
$f_2=f\left(\frac{v}{v+v_s}\right)=f{(1+\frac{v_s}{v})}^{-1}=f(1-\frac{v_s}{v})\therefore \left(\frac{f_2-f}{f}\right)\times 100=-\frac{v_s}{v}\times 100=-10$
[from Eq.(i)]
In the first case, observed frequency increases by $10\%$ while in the second case, observed frequency decreases by $10\%$