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 close to:

• A. 6%
• B. 12%
• C. 18%
• D. 24%

Answer 1

The correct option is B 12%

A train is moving on a straight track with speed vs = 20 m/s. It is blowing its whistle at the frequency of $v_0=1000Hz$.
A person is standing near the track.
As the train approaches him, the frequency of the whistle as heard by him is $v_1=v_0\frac{v}{v-v_s}$ (where v = 320 m/s is the speed of sound in air)
$\therefore v_1=1000\times \frac{320}{(320-20)}=1000\times \frac{320}{300}=1066.67Hz$.
As the train goes away from him, the frequency of the whistle as heard by him is $v_2=v_0\frac{v}{v+v_s}$
$v_2=1000\times \frac{320}{(320+20)}=1000\times \frac{320}{340}=941.17Hz$
Percentage change in frequency heard by the person standing near the track as the train passes him is $\frac{v_1-v_2}{v_0}\times 100=\frac{1066.67-941.17}{1000}\times 100=\frac{125.5}{10}=12.55\%\approx 12\%$