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 \(380\ \text{Hz}\), find the original frequency of tuning fork \(A\). Given: \[ f_B = 380\ \text{Hz} \]

Hint:

Adding wax to a tuning fork always decreases its frequency due to increase in effective mass.

Answer 1

Correct Answer: 384

Given that the number of beats per second is the absolute difference between the frequencies of the two tuning forks, we can interpret the given information as follows: The beat frequency before loading wax is \( \frac{8}{2} = 4\ \text{Hz} \). Since this is the difference between the frequencies of forks \(A\) and \(B\), we have two possibilities for fork \(A\):

\(f_A = f_B \pm 4\). Given \(f_B = 380\ \text{Hz}\), fork \(A\) initially could have a frequency of:

• \(f_A = 380 + 4 = 384\ \text{Hz}\)
• or \(f_A = 380 - 4 = 376\ \text{Hz}\)

When fork \(A\) is loaded with wax, its frequency decreases, and the beat frequency becomes \( \frac{4}{2} = 2\ \text{Hz} \). Therefore, after loading wax, \(f_A^\prime\) is such that:

\(|f_A^\prime - 380| = 2\).

This gives us two possibilities for \(f_A^\prime\):

• \(f_A^\prime = 380 + 2 = 382\ \text{Hz}\)
• or \(f_A^\prime = 380 - 2 = 378\ \text{Hz}\)

We know the frequency of fork \(A\) must decrease upon loading with wax. If \(f_A = 384\ \text{Hz}\), loading wax could result in:

\(f_A \rightarrow 382\ \text{Hz}\) (matches decreased frequency requirement).

If \(f_A = 376\ \text{Hz}\), loading wax would result in:

\(f_A \rightarrow 378\ \text{Hz}\) (contradicts decreased frequency requirement).

Thus, the original frequency of tuning fork \(A\) must be:

384 Hz

This frequency, 384 Hz, fits within the expected range [384, 384].