Waves - Beats

A tuning fork $P$ of unknown frequency gives $6\text{ beats/s}$ with another tuning fork $Q$ of frequency $256\text{ Hz}$. On loading $P$ with a little wax, the number of beats per second remains $6$. Find the original frequency of tuning fork $P$.

The original frequency of $P$ is $262\text{ Hz}$.

The possible frequencies for $P$ are $f_P = 256 \pm 6$, which gives $262\text{ Hz}$ or $250\text{ Hz}$. When $P$ is loaded with wax, its frequency $f_P$ decreases. Case 1: If $f_P = 250\text{ Hz}$, decreasing it makes the difference with $256\text{ Hz}$ greater than $6$. Case 2: If $f_P = 262\text{ Hz}$, decreasing it to $250\text{ Hz}$ would make the difference $|250 - 256| = 6$. Since the beat frequency remains $6$, the original frequency must have been higher than $256$. Thus, original $f_P = 262\text{ Hz}$.

Two sound sources produce waves given by $y_1 = 5 \sin(100\pi t)$ and $y_2 = 5 \sin(108\pi t)$. Calculate the beat frequency and the time interval between successive maximum intensities.

$f_{beat} = 4\text{ Hz}$, $T_{beat} = 0.25\text{ s}$

Comparing with $y = A \sin(2\pi f t)$, we get: $2\pi f_1 = 100\pi \implies f_1 = 50\text{ Hz}$ and $2\pi f_2 = 108\pi \implies f_2 = 54\text{ Hz}$. The beat frequency is $f_{beat} = |f_2 - f_1| = |54 - 50| = 4\text{ Hz}$. The time interval between successive maxima is $T = \frac{1}{f_{beat}} = \frac{1}{4} = 0.25\text{ s}$.