Question

The frequency of the whistle of an engine appears to be ${\left(\frac{5}{6}\right)}^{th}$ of the original frequency , when it passes a stationary observer. If the speed of sound in air is $350\text{m/s}$, then the speed of the engine is:

[Assume, medium is stationary and the original frequency of the whistle is measured when the source and observer are at rest]

• A. $35\text{m/s}$
• B. $70\text{m/s}$
• C. $105\text{m/s}$
• D. $140\text{m/s}$

Answer 1

The correct option is B $70\text{m/s}$

Given,
Speed of sound in stationary air medium, $v_s=350\text{m/s}$

Let the original frequency of the whistle be $f_0$ and the speed of the engine be $v$ when it passes a stationary observer.

Frequency heard by the stationary observer, $f_{app}=\frac{5}{6}{f}_0$
$\Rightarrow f_{app}<f_0$

So, we can say that, source of sound is moving away from the observer.

Now, the apparent frequency heard by the observer, $f_{app}=\left(\frac{v}{v+v_s}\right)f_0$

$\Rightarrow \frac{5}{6}{f}_0=f_0\left[\frac{350}{350+v}\right]$

$\Rightarrow 5\times 350+5v=6\times 350\Rightarrow v=\frac{350}{5}$

$\Rightarrow v=70\text{m/s}$