Question

A train is moving on a straight track with speed $20m{s}^{-1}$. It is blowing its whistle at the frequency of $1000Hz$. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound $=320m{s}^{-1}$) close to :

• A. $6\%$
• B. $12\%$
• C. $18\%$
• D. $24\%$

Answer 1

Answer: (B)

$12\%$

The frequency heard by the listener when train is moving towards the listener is given by,
$f^{\prime }=\frac{v}{v-v_s}f\Rightarrow f^{\prime }=\frac{1000v}{v-20}$
The frequency heard by the listener when train is moving away from the listener is given by,
$f^{\prime \prime }=\frac{v}{v+v_s}f\Rightarrow f^{\prime \prime }=\frac{1000v}{v+20}$
And percentage change in frequency will be given by,

Percentage change $=\frac{f^{\prime }-f^{\prime \prime }}{f}\times 100=\frac{\frac{1000v}{v-20}-\frac{1000v}{v+20}}{1000}\times 100=\frac{40v}{(v+20)(v-20)}\times 100=\frac{40\times 320}{(320+20)(320-20)}\times 100$

Percentage change $=\frac{40\times 320}{340\times 300}\times 100=12.459$%$\approx 12$%