Two tuning forks $A$ and $B$ are sounded together giving rise to 8 beats in 2 s . When fork $A$ is l

Question

Two tuning forks $A$ and $B$ are sounded together giving rise to 8 beats in 2 s . When fork $A$ is loaded with wax, the beat frequency is reduced to 4 beats in 2 s . If the original frequency of tuning fork $B$ is 380 Hz then original frequency of tuning fork $A$ is $\_\_\_\_$ Hz.

Professor PC Thomas Classes · Answer

The initial beat frequency is $n_1 = \frac{8 	ext{ beats}}{2 	ext{ s}} = 4 	ext{ Hz}$. Thus, the original frequency of A could be $f_A = f_B \pm 4 = 380 \pm 4$, meaning $f_A$ is either $384 	ext{ Hz}$ or $376 	ext{ Hz}$. When tuning fork A is loaded with wax, its mass increases, causing its frequency to decrease. The new beat frequency is $n_2 = \frac{4 	ext{ beats}}{2 	ext{ s}} = 2 	ext{ Hz}$. If $f_A$ was 376 Hz, decreasing it would widen the gap from 380 Hz, increasing the beat frequency. If $f_A$ was 384 Hz, decreasing it moves it closer to 380 Hz, dropping the beat frequency to 2 Hz. Since the beat frequency reduced, the original frequency of A must be $384 	ext{ Hz}$.