Question

Suppose that the speed of sound in air at a given temperature is 400 m/sec. An engine blows a whistle at 1200 Hz frequency. It is approaching an observer at the speed of 100 m/sec. What is the apparent frequency as heard by the observer [MP PMT 1991]

• A. 600 Hz
• B. 1200 Hz
• C. 1500 Hz
• D. 1600 Hz

Answer 1

The correct option is D 1600 Hz

We know that the apparent frequency due to doppler effect is given by,

$f^{\prime }=f\left(\frac{v\pm v_o}{v\pm v_s}\right)$

Here, velocity of sound in the medium, $v=400m/s$. The observer is at rest, and hence $v_o=0$. The velocity of the source, $v_s=100m/s$ and the frequency of the source, $f=1200Hz$.

The source is coming towards the observer and therefore the sign used here will be $-$.

Using given values we have,
$f^{\prime }=1200\left(\frac{400}{400-100}\right)$

or, $f^{\prime }=1600Hz$, which is our required answer.

The correct answer is option (C).