When a train is approaching the observer, the frequency of the whistle is 100 cps while when it has passed the observer, it is 50 cps. Calculate the frequency when the observer' moves with the trail.

Explanation

According to Doppler effect in case of approachign of source, `f. = f v/(v -v_s) , i.e., 100 = (fv)/(v -v_s)`
`i.e., v -v_s = (fv)/(100)`, while in case of recession of source.
`f. = f v/(v +v_s) i.e., 50 = (fv)/(v +v_s) , v +v_s =(fv)/50`
So adding the tow eqns., `2V =(3fv)/100`
`i.e., f =200/3 = 66.67 Hz`