Question

Two sources A and B are producing notes of frequency 680 Hz. A detector moves from A to B with a constant velocity 2 m/s. Find number of beats heard by the detector per second, if velocity of sound v is 340 m/s.

Answer 1

The frequency of source $A$ received by the detector

$f_1=\left(\frac{v-v_0}{v}\right)f....\left(i\right)$

The frequency of source B received by the observer

$f_2=\left(\frac{v+v_0}{v}\right)f....\left(ii\right)$

Now number of beats heard by the detector per second i.e.,

Beat frequency $=|f_1-f_2|$
$=[\left(\frac{v+v_0}{v}\right)-\left(\frac{v-v_0}{v}\right)]$
$=\frac{2v_{0}f}{v}$
$=\frac{2\times 2\times 680}{340}$
$=8Hz$