Question

Two tuning forks produce $4$ beats per second when they are sounded together. Now, both the forks are moved towards the observer at same speed $\left(u\right)$. The beat frequency now becomes $5\text{Hz}$. If the observer also begins to run with speed $u$ towards both the forks, what beat frequency will he hear now?

• A. $4\text{Hz}$
• B. $8\text{Hz}$
• C. $6\text{Hz}$
• D. $10\text{Hz}$

Answer 1

The correct option is C $6\text{Hz}$

Let us assume frequencies of two forks are $f$ and $f+4$ and let $v$ be the velocity of sound waves in a stationary medium.

When forks are moving towards the observer, the frequencies received by the observer are given by

$f_1=f\left(\frac{v}{v-u}\right),f_2=(f+4)\left(\frac{v}{v-u}\right)$

$\therefore \Delta f=|f_2-f_1|=4\left(\frac{v}{v-u}\right)$

From the data given in the question,

$5=4\left(\frac{v}{v-u}\right)\Rightarrow 5v-5u=4v\Rightarrow u=\frac{v}{5}....\left(1\right)$

When observer also begins to run, the frequencies received by him will be

$f_1^{{}^{\prime }}=f\left(\frac{v+u}{v-u}\right),f_2^{{}^{\prime }}=(f+4)\left(\frac{v+u}{v-u}\right)$

$\therefore \Delta f^{\prime }=|f_2^{{}^{\prime }}-f_1^{{}^{\prime }}|=4\left(\frac{v+u}{v-u}\right)$

Substituting $\left(1\right)$ in the above equation, we get

$\Delta f^{\prime }=4\left(\frac{v+\frac{v}{5}}{v-\frac{v}{5}}\right)=4\times \frac{6}{4}=6\text{Hz}$
will be the beat frequency heard by him.