Oscillations and Waves - Beats

Two tuning forks $A$ and $B$ produce $4$ beats per second. The frequency of $A$ is $256 \text{ Hz}$. When fork $B$ is loaded with a little wax, the number of beats increases to $6$ per second. Find the frequency of fork $B$.

1. Known frequency $f_A = 256 \text{ Hz}$.
2. Possible frequencies for $B$ are $f_B = f_A \pm n = 256 \pm 4$, which gives $f_B = 260 \text{ Hz}$ or $f_B = 252 \text{ Hz}$.
3. Loading $B$ with wax decreases $f_B$.
4. Case 1: If $f_B = 260 \text{ Hz}$, decreasing it (e.g., to $258 \text{ Hz}$) would result in $|256 - 258| = 2$ beats (beats decrease). This contradicts the problem.
5. Case 2: If $f_B = 252 \text{ Hz}$, decreasing it (e.g., to $250 \text{ Hz}$) would result in $|256 - 250| = 6$ beats. This matches the problem.
6. Therefore, the frequency of $B$ is $252 \text{ Hz}$.

We use the principle that loading a fork reduces its frequency and then compare the new beat count with the original to determine which initial frequency was correct.

Two sound waves are given by $y_1 = 5 \sin(100 \pi t)$ and $y_2 = 5 \sin(104 \pi t)$. Calculate the beat frequency and the number of beats heard in $5$ seconds.

1. Compare $y = a \sin(2\pi f t)$ with the given equations.
2. For $y_1$, $2\pi f_1 = 100\pi \implies f_1 = 50 \text{ Hz}$.
3. For $y_2$, $2\pi f_2 = 104\pi \implies f_2 = 52 \text{ Hz}$.
4. Beat Frequency $n = |f_2 - f_1| = |52 - 50| = 2 \text{ beats/s}$.
5. Total beats in $5$ seconds $= n \times t = 2 \times 5 = 10$ beats.

The beat frequency is the absolute difference between the individual frequencies of the two interfering waves.