Question

Two trains, each moving with a velocity of $30m{s}^{-1}$, cross each other. One of the trains gives a whistle whose frequency is $600Hz$. If the speed of sound is $330m{s}^{-1}$, the apparent frequency for passengers sitting in the other train before crossing would be

• A. $600Hz$
• B. $630Hz$
• C. $920Hz$
• D. $720Hz$

Answer 1

The correct option is D

$720Hz$

Step 1: Given data

Speed of sound, $v=330m{s}^{-1}$

The velocity of both the trains, $v_0=v_s=30m{s}^{-1}$

Frequency of the whistle from the train, $f=600Hz$

Step 2: To find

The apparent frequency for passengers sitting in the other train.

Step 3: Formula used

Apparent frequency,

$f\text{'}=f\left(\frac{v+v_0}{v-v_s}\right)$

Step 4: Calculation

Apparent frequency,

$$\begin{gathered} f\text{'}=f\left(\frac{v+v_0}{v-v_s}\right) \\ \Rightarrow f\text{'}=600\times \left(\frac{330+30}{330-30}\right) \\ \Rightarrow f\text{'}=600\times \left(\frac{360}{300}\right) \\ \Rightarrow f\text{'}=720Hz \end{gathered}$$

The apparent frequency for passengers sitting in the other train is $720Hz$.

Option D is correct.