91Ó°ÊÓ

1. Speed of each train is V_t = 30.5m/s
2. Frequency of whistle is f = 500Hz
3. Speed of the sound is v = 343m/s.

We can use the concept of the Doppler Effect. We can use the equation of the Doppler Effect for motion of source and listener. When there is air moving between trains, we find their relative speed, and then we can find the frequency heard by the listener.

Formula:

The frequency received by the observer according to Doppler’s Effect,

$\mathit{f}\mathbf{\text{'}}\mathbf{=}\left(\frac{VÂ\pm V_l}{VÂ\pm V_s}\right)\mathit{f}$ …(¾±)

Here boththesource and the listener are moving towards each other, so using equation (i), we can write the frequency as:

$\begin{array}{rcl}f\text{'}& =& 500Hz×\left(\frac{343m/s+30.5m/s}{343m/s-30.5m/s}\right)\\ & =& 598Hz\end{array}$

Hence, the frequency heard at the other train is 598Hz.

In this frame of reference, the air seems still because both listener and air are moving with the same speed, so their relative speed will be zero.

At the source side, the speed of the train and air is opposite, so their relative speed will be 2(30.5m/s) = 61m/s We can write the frequency using equation (i) as:

$\begin{array}{rcl}f\text{'}& =& 500Hz×\left(\frac{343m/s+0}{343m/s-61m/s}\right)\\ & =& 608Hz\end{array}$

Hence, the required frequency when wind is blowing at 30.5m/s is 608Hz.

Here, the relative speed of the source and air will be zero while the relative speed of the listener and air will be 61m/s. Hence, the frequency can be given using equation (i) as:

$\begin{array}{rcl}f\text{'}& =& 500Hz×\left(\frac{343m/s+61m/s}{343m/s-0}\right)\\ & =& 589Hz\end{array}$

Hence, the required frequency is 589Hz.