Question

A man standing on a platform observes that the frequency of the sound of a whistle emitted by a moving train is less by $140\text{Hz}$ that the frequency emitted when the train is stationary. If the velocity of sound in air is $330\text{m/s}$ and the speed of the train is $70\text{m/s}$, the frequency of the whistle is

• A. $571\text{Hz}$
• B. $800\text{Hz}$
• C. $400\text{Hz}$
• D. $260\text{Hz}$

Answer 1

The correct option is B $800\text{Hz}$

Let $f_0$ be the frequency of sound heard by the observer, when both the source of sound and observer are at rest.
Let $v$ be the velocity of sound in the stationary medium and $v_s$ be the velocity of the source of sound.

Since the frequency of whistle heard by the observer is less, we can say that the train is moving away from observer.
When a source is moving away from a stationary observer, the frequency heard by the observer is given by
$f_{app}=f_0\left(\frac{v}{v+v_s}\right)......\left(1\right)$

From the data given in the question,
Decrease in frequency $f_0-f_{app}=140\text{Hz}$

Using $\left(1\right)$ in the above equation,

$f_0\left(\frac{v_s}{v+v_s}\right)=140$

Given,
Velocity of sound in stationary medium $v=330\text{m/s}$

Velocity of source of sound $v_s=70\text{m/s}$

Substituting these in the above equation, we get

$f_0\left(\frac{70}{330+70}\right)=140$

$\Rightarrow f_0=\frac{140\times 400}{70}=800\text{Hz}$