A violin player riding on a slow train plays a 440Hz note. Anothe… - Vidyadip

Question

A violin player riding on a slow train plays a 440Hz 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 = 340m/s.

Calculate the speed of the train.
What beat frequency is heard by the player in the train?

✓

Answer

Given,

Frequency of violins f₀= 440Hz

Speed of sound in air v = 340ms^-1

Let the velocity of the train (sources) be v_s.

Beat heard by the standing man = 4

$\therefore\ \text{Frequency (f}_1)=440+4$

$\text{f}_1=444\text{Hz or 436 Hz}$

Now,

$\text{f}_1=\Big(\frac{340}{340-\text{v}_\text{s}}\Big)\times\text{f}_0$

On substituting the values,

We have,

$444=\Big(\frac{340+0}{340-\text{v}_\text{s}}\Big)\times440$

$\Rightarrow444(340-\text{v}_\text{s})=440\times340$

$\Rightarrow340\times(444-440)=440\times\text{v}_\text{s}$

$\Rightarrow340\times4=440\times\text{v}_\text{s}$

$\Rightarrow\text{v}_\text{s}=3.09\text{m/ s}=11\text{km/ h}$

The sitting man will listen to fewer than 4 beats/-s.

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