Question

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

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

Answer 1

The correct option is B $12\%$

Here, the listener is at rest (on the platform). Using Doppler's effect formula,
$f_{app}=f_0\left(\frac{C}{C\pm V_s}\right)$
where $C=320\text{m/s}=$ speed of sound
& $V_s=$ speed of the train $=20\text{m/s}$

Apparent frequency heard by the person when the train is coming towards him,
$f_1=\left(\frac{C}{C-V_s}\right)f_0=\left(\frac{320}{320-20}\right)1000$

Similarly, apparent frequency heard when train is moving away,
$f_2=\left(\frac{C}{C+V_s}\right)f_0=\left(\frac{320}{320+20}\right)1000$

Therefore,
$\Delta f=f_1-f_2=\left(\frac{2C{V}_s}{C^2-V_s^2}\right)f_0$
and $\frac{\Delta f}{f_0}\times 100=\left(\frac{2C{V}_s}{C^2-V_s^2}\right)\times 100$

$=\frac{2\times 320\times 20}{{320}^2-{20}^2}\times 100=\frac{2\times 32\times 20}{3\times 34}$

$=12.54\%\approx 12\%$