Question

Two sources A and B are sounding notes of frequency $680\text{Hz}$. A listener moves from A to B with a constant velocity $u$. If the speed of sound is $340\text{m/s}$, what must be the value of $u$ so that he hears $10$ beats per second?

• A. $2.0\text{m/s}$
• B. $2.5\text{m/s}$
• C. $30\text{m/s}$
• D. $3.5\text{m/s}$

Answer 1

The correct option is B $2.5\text{m/s}$

Apparent frequency due to source A is
$n^{\prime }=\frac{v-u}{v}\times n$
Apparent frequency due to source B is
$n^{\prime \prime }=\frac{v+u}{v}\times n$

$\therefore$ beat frequency $\Delta n=n^{\prime \prime }-n^{\prime }=\frac{2u}{v}\times n=10$ (given)
$\therefore u=\frac{10v}{2n}=\frac{10\times 340}{2\times 680}=2.5\text{m/s}$

Why this question? Note: When two sound waves with different frequencies, say $n_1$ and $n_2$ superimpose, the no. of beats produced is given by $|n_1-n_2|$.