Question

A person observes two trains one of them is coming towards with speed of $4\text{m/sec}$ and another is going away with same speed. If the two trains blowing a whistle with frequency $240\text{Hz}$. The beat frequency heard by stationary person will be (speed of sound in air $=320\text{m/sec}$.)

• A. zero
• B. $3$
• C. $6$
• D. $12$

Answer 1

The correct option is C $6$

Let $v_{s_1}$ and $v_{s_2}$ be the velocities of the two sources then the frequencies heard by the observer from each of the source is given by
$n_1=n\left(\frac{v}{v-v_{s_1}}\right)=240\left(\frac{320}{316}\right)=243\text{Hz}$
$n_2=n\left(\frac{v}{v+v_{s_2}}\right)=240\left(\frac{320}{324}\right)=237\text{Hz}$
$\text{beats}=n_1-n_2=(243-237)\text{Hz}=6\text{Hz}$