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 Closed by and the Train Approaches the Person - Physics

Sum

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 closed 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−1. (a) Calculate the speed of the train. (b) What beat frequency is heard by the player in the train?

Solution

Given:
Frequency of violins $f_0$= 440 Hz
Speed of sound in air v = 340 ms−1
Let the velocity of the train (sources) be vs.

(a) Beat heard by the standing man = 4
∴ frequency $\left( f_1 \right)$= 440 + 4
= 444 Hz or 436 Hz

Now,

$f_1 = \left( \frac{340}{340 - v_s} \right) \times f_0$

On substituting the values, we have:

$444 = \left( \frac{340 + 0}{340 - v_s} \right) \times 440$

$\Rightarrow 444\left( 340 - v_s \right) = 440 \times 340$

$\Rightarrow 340 \times \left( 444 - 440 \right) = 440 \times v_s$

$\Rightarrow 340 \times 4 = 440 \times v_s$

$\Rightarrow v_s = 3 . 09 \text { m/s }= 11 \text { km/h }$

(b) The sitting man will listen to fewer than 4 beats/s.

Concept: Speed of Wave Motion
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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 16 Sound Waves
Q 69 | Page 356