Question

Two tuning forks \(A\) and \(B\) produce \(8\) beats in \(2\,\text{s}\) when sounded together. The frequency of tuning fork \(B\) is \(380\,\text{Hz}\). If tuning fork \(A\) is loaded with some wax, then they produce \(4\) beats in \(2\,\text{s}\). Find the original frequency of tuning fork \(A\).

• A. \(384\)
• B. \(388\)
• C. \(380\)
• D. \(392\)

Hint:

Adding wax to a tuning fork {always decreases its frequency}. Use this fact to eliminate incorrect solutions in beat problems.

Answer 1

The Correct Option is A

To determine the original frequency of tuning fork \(A\), we use the concept of beats produced when two tuning forks of slightly different frequencies are sounded together.

Beats are produced due to the superposition of sound waves of different frequencies, and the beat frequency is equal to the absolute difference between their frequencies.

Given:

• Frequency of tuning fork \(B = 380\,\text{Hz}\)
• Number of beats produced = 8 in 2\,\text{s}
• After loading tuning fork \(A\) with wax, beats reduce to 4 in 2\,\text{s}

Step-by-Step Solution:

1. Initial beat frequency:

   |f_A - f_B| = \dfrac{8}{2} = 4\,\text{Hz}

2. Possible values of frequency of tuning fork \(A\):

   f_A = 380 + 4 = 384\,\text{Hz}

   f_A = 380 - 4 = 376\,\text{Hz}

3. After loading tuning fork \(A\) with wax, the beat frequency becomes:

   |f'_A - f_B| = \dfrac{4}{2} = 2\,\text{Hz}

4. Loading a tuning fork with wax decreases its frequency. If the original frequency of \(A\) were \(384\,\text{Hz}\), reducing it would bring the frequency closer to \(380\,\text{Hz}\), giving a smaller beat frequency of \(2\,\text{Hz}\).
5. If the original frequency were \(376\,\text{Hz}\), lowering it further would increase the difference from \(380\,\text{Hz}\), which contradicts the observed decrease in beat frequency.

Therefore, the original frequency of tuning fork \(A\) is:

\(\boxed{384\,\text{Hz}}\)