Question

A train moves towards a stationary observer with a speed $34\text{m/s}$. The train sounds a whistle and its frequency registered by the observer is $f_1$. If the speed of the train is reduced to $17\text{m/s}$, the frequency registered is $f_2$. If the speed of sound is $340\text{m/s}$ , then the ratio $\frac{f_1}{f_2}$ is

[Assume, medium is stationary ]

• A. $\frac{18}{19}$
• B. $\frac{1}{2}$
• C. $2$
• D. $\frac{19}{18}$

Answer 1

The correct option is D $\frac{19}{18}$

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 $v_s=34\text{m/s}$,

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

$f_1=f_0\left(\frac{340}{340-34}\right)=\frac{340}{306}{f}_0.....\left(2\right)$

Case -2:

When the speed of train is $v_s=17\text{m/s}$
Using $\left(1\right)$, we get
$f_2=f_0\left(\frac{340}{340-17}\right)=\frac{340}{323}{f}_0......\left(3\right)$

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

$\frac{f_1}{f_2}=\frac{323}{306}\Rightarrow \frac{f_1}{f_2}=\frac{19}{18}$