Question

A source moves towards a stationary observer with speed $40\text{m/s}$. The source sounds a whistle and its frequency registered by the observer is $n_1$. If the source's speed is reduced to $20\text{m/s}$, the frequency registered is $n_2$. If the speed of sound is $340\text{m/s}$, then the ratio $\frac{f_1}{f_2}$is

[Assume, medium is stationary]

• A. $\frac{5}{4}$
• B. $\frac{15}{16}$
• C. $\frac{16}{15}$
• D. $\frac{4}{5}$

Answer 1

The correct option is C $\frac{16}{15}$

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.

When a source is moving towards 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,

Case -1 :

When the speed of train is $v_s=40\text{m/s}$

Using $\left(1\right)$, we get

$f_1=f_0\left(\frac{340}{340-40}\right)=\frac{340}{300}{f}_0.....\left(2\right)$

Case -2:

When the speed of train is $v_s=20\text{m/s}$
Using $\left(1\right)$, we get
$f_2=f_0\left(\frac{340}{340-20}\right)=\frac{340}{320}{f}_0......\left(3\right)$

From $\left(2\right)$ and $\left(3\right)$, we get

$\frac{f_1}{f_2}=\frac{320}{300}\Rightarrow \frac{f_1}{f_2}=\frac{16}{15}$.