Question

A violin player riding on a slow train plays a 440 Hz note. Another violin player standing near the track plays the same note. When the two are close by and the train approaches the person on the ground, he hears 4.0 beats per second. The speed of sound in air = 340 m/s. (a) Calculate the speed of the train. (b) What beat frequency is heard by the player in the train ?

Answer 1

Here, $\theta =ta{n}^{-1}\left(\frac{0.5}{2.4}\right)={22}^{\circ }$

So the velocity of the source will be 'v cos ${\theta }^{\prime }$ when heard by the observer.

So the apparent frequency received by the man from train B.

$f^1=\left(\frac{340+0+0}{340-vcos{22}^{\circ }}\right)500$

And the apparent frequency heard by the man from Train C,

$f^{11}=\left(\frac{340+0+0}{340+vcos{22}^{\circ }}\right)\times 500$

=476 Hz