Two tuning forks with natural frequencies of 340 Hz each move relative to a stationary observer. One

Question

Two tuning forks with natural frequencies of 340 Hz each move relative to a stationary observer. One fork moves away from the observer, while the other moves towards him at the same speed. The observer hears beats of frequency 3 Hz . Find the speed of the tuning fork. (Speed of sound $=340 \mathrm{~m} / \mathrm{s}$ )\

Professor PC Thomas Classes · Answer

According to Doppler's effect, the apparent frequencies heard by the stationary observer will be different from the approaching and receding forks.\ For fork moving towards observer, $$ f^{\prime}=\frac{v}{\left(v-v_{S}ight)} f $$ where $v_{S}=$ velocity of fork.\ For fork moving away from observer, $$ f^{\prime \prime}=\frac{v}{v+v_{S}} f $$ where $v=$ speed of sound.\ $	herefore \quad$ Beat frequency $=f^{\prime}-f^{\prime \prime}$\ or $3=\frac{v f}{v-v_{S}}-\frac{v f}{v+v_{S}}$\ or $3=v f\left[\frac{\left(v+v_{S}ight)-\left(v-v_{S}ight)}{\left(v-v_{S}ight)\left(v+v_{S}ight)}ight]$\ or $3=\frac{v f 	imes 2 v_{S}}{v^{2}-v_{S}^{2}}$\ or $3=\frac{v f 	imes 2 v_{S}}{v^{2}}$ as $v_{S} \ll v$\ or $\quad v_{S}=\frac{3 v^{2}}{2 v f}=\frac{3 v}{2 f}$ or $v_{S}=\frac{3 	imes 340}{2 	imes 340}$ or $\quad v_{S}=1.5 \frac{\mathrm{~m}}{\mathrm{~s}}$\ $	herefore \quad$ Speed of tuning fork $=1.5 \frac{\mathrm{metre}}{\mathrm{sec} 	ext {. }}$.\