Question

A train is moving on a straight track with speed $40\text{m/s}$. It is blowing its whistle at the frequency of $1200\text{Hz}$. 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 $v=340\text{m/s})$]

• A. $15\%$
• B. $17\%$
• C. $21\%$
• D. $25\%$

Answer 1

The correct option is C $21\%$

Since observer is stationary and source is moving with velocity $40\text{m/s}$
$\left(v_o\right)=0\text{m/s};\left(v_s\right)=40\text{m/s}$

Velocity of sound $\left(v\right)=340\text{m/s}$

Initially, when the train is approaching the person,

$f_1=f_0\left(\frac{v}{v-v_s}\right)$

From the data given in the question,

$f_1=1200\times \left(\frac{340}{340-40}\right)$

$\Rightarrow f_1=1200\times \frac{340}{300}=1360\text{Hz}$

After the train passes the person and starts moving away from him,

$f_2=f_0\left(\frac{v}{v+v_s}\right)$

From the data given in the question,

$f_2=1200\times \frac{340}{380}=1073.68\approx 1074\text{Hz}$

So $\%$ change in frequency $=\frac{f_1-f_2}{f_1}\times 100=\frac{1360-1074}{1360}\times 100=21\%$