Question

A trumpet player on a moving railroad flatcar moves toward a second trumpet player standing alongside the track both play a 490 Hz note. The sound waves heard by a stationary observer between the two players have a beat frequency of 2.0 beats/s. What is the flatcar's speed? Take the speed of sound to be 343 m/s.

Answer 1

Answer:

The value is $v_s = 1.394 \ m/s$

Explanation:

From the question we are told that

The frequency of the second player is $f_2 = 490 \ Hz$

The beat frequency is $f_b = 2.0 \ Hz$

The speed of sound is $v_s = 343 \ m/s$

Generally the frequency of the note played by the first player is mathematically represented as

$f_1 = f_2 + f_b$

=> $f_1 = 490 + 2.0$

=> $f_1 = 492 Hz$

From the relation of Doppler Shift we have that

$f_1 = \frac{ f_2 (v+ v_o )}{v-v_s }$

Here $v_o\ is\ the\ velocity\ of\ the\ observer\ with\ value\ 0 \ m/s$

So

$492 = \frac{ 490 (343+0 )}{343 -v_s }$

=> $v_s = 1.394 \ m/s$