Question

Two identical tuning forks vibrating at the same frequency $256Hz$ are kept fixed at some distance apart. A listener runs between the forks at a speed of $3.0m/s$ so that the approaches one tuning fork and recedes from the other figure. Then the beat frequency observed by the listener. Speed of sound in air $=332m/s$.

Answer 1

Speed of sound in air$V=332m/s$,

The speed of the observer, $u=3.0m/s$.

The frequency of the sources $f_0=256Hz.$

The sources are stationary and observer moving, hence the apparent frequency heard by the observer from the front tuning fork (approaching)

$f^{\prime }=\frac{V+u}{V}\times f_0=\frac{(332+3)}{332}\times 256=258Hz$

and the apparent frequency heard by the observer from the receding source

$f^{\prime \prime }=\frac{V-u}{V}\times f_0=\frac{(332-3)}{332}\times 256=246Hz$

Hence the beat frequency observed by the listener is,

$=f^{\prime }-f^{\prime \prime }$

$=258-246$

$=12Hz$