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

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 $= 340m/s.$

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

✓

Answer

Given, Frequency of violins $f_{0 }= 440\ Hz$ 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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