Question

A tuning fork of unknown frequency sounded with a tuning fork of frequency 256 Hz produces 4 beats per second. If a small quantity of wax is fixed on first fork so that it produces 3 beats per second with tuning fork, what will be the frequency of first fork (in Hz)?

• A. 260
• B. 252
• C. 256
• D. 280

Hint:

Remember the effects of modifying a tuning fork: - **Adding mass (waxing):** Decreases frequency. - **Removing mass (filing):** Increases frequency.

Answer 1

The Correct Option is A

Step 1: Determine the possible initial frequencies of the unknown fork. Let the unknown frequency be \(f_1\). The reference frequency is \(f_2 = 256\) Hz. The beat frequency is \(f_{beat} = |f_1 - f_2| = 4\) Hz. This gives two possibilities for \(f_1\): - \(f_1 = 256 + 4 = 260\) Hz - \(f_1 = 256 - 4 = 252\) Hz
Step 2: Analyze the effect of adding wax. Adding wax to a tuning fork increases its mass, which causes its frequency to decrease. Let the new frequency of the first fork be \(f_1'\), where \(f_1'<f_1\).
Step 3: Test the two possibilities with the new information. The new beat frequency is 3 Hz. - Case A: Assume initial \(f_1 = 260\) Hz. When wax is added, its frequency \(f_1'\) will be slightly less than 260 Hz. The new beat frequency is \(|f_1' - 256|\). As \(f_1'\) moves from 260 towards 256, the difference decreases. It is possible for the beat frequency to become 3 Hz. This is consistent. - Case B: Assume initial \(f_1 = 252\) Hz. When wax is added, its frequency \(f_1'\) will be slightly less than 252 Hz. The new beat frequency is \(|f_1' - 256| = 256 - f_1'\). As \(f_1'\) decreases from 252, it moves further away from 256, so the difference will increase. The beat frequency would become greater than 4 Hz, not 3 Hz. This contradicts the observation.
Step 4: Conclude the initial frequency. Only the first case is consistent with the experimental result. Therefore, the original frequency of the first fork was 260 Hz.