When a tuning fork of frequency 341 is sounded with another tuning fork, six beats per second are heard. When the second tuning fork is loaded with wax and sounded with the first fork, the number of beats is two per second. The natural frequency of the second tuning fork is

Options

A. 334
B. 339
C. 343
D. 347

Answer

To find the natural frequency of the second tuning fork, we can use the information given about the beat frequencies when the tuning forks are sounded together.

### Step-by-Step Solution:

1. **Identify the Frequencies and Beat Frequencies:**
- Let the frequency of the first tuning fork (Tuning Fork 1) be \( f_1 = 341 \, \text{Hz} \).
- The beat frequency when the second tuning fork (Tuning Fork 2) is sounded with the first is \( f_b = 6 \, \text{Hz} \).
- When Tuning Fork 2 is loaded with wax, the new beat frequency is \( f_b' = 2 \, \text{Hz} \).

2. **Determine Possible Frequencies for Tuning Fork 2:**
- The beat frequency is given by the absolute difference between the frequencies of the two tuning forks:
\[ f_b = |f_1 - f_2| \]
- Therefore, we have two cases for the natural frequency of Tuning Fork 2 (\( f_2 \)):
- Case 1: \( f_2 = f_1 + 6 = 341 + 6 = 347 \, \text{Hz} \)
- Case 2: \( f_2 = f_1 - 6 = 341 - 6 = 335 \, \text{Hz} \)

3. **Analyze the Effect of Loading with Wax:**
- When the second tuning fork is loaded with wax, its frequency decreases. Therefore, we need to evaluate the two possible frequencies from the previous step:
- If \( f_2 = 347 \, \text{Hz} \), after loading, the frequency becomes less than 347 Hz.
- If \( f_2 = 335 \, \text{Hz} \), after loading, the frequency becomes less than 335 Hz.

4. **Calculate New Beat Frequencies:**
- For \( f_2 = 347 \, \text{Hz} \):
\[ f_b' = |341 - (347 - x)| = |341 - 347 + x| = |x - 6| \]
Setting \( |x - 6| = 2 \) gives \( x = 4 \) or \( x = 8 \). This means the frequency could be \( 343 \, \text{Hz} \) or \( 349 \, \text{Hz} \), but since it must be less than 347 Hz, we discard this case.

- For \( f_2 = 335 \, \text{Hz} \):
\[ f_b' = |341 - (335 - x)| = |341 - 335 + x| = |x + 6| \]
Setting \( |x + 6| = 2 \) gives \( x = -4 \) or \( x = -8 \). This means the frequency could be \( 331 \, \text{Hz} \) or \( 337 \, \text{Hz} \), but since it must be less than 335 Hz, we discard this case.

5. **Conclusion:**
- The only feasible case is when the natural frequency of Tuning Fork 2 is \( f_2 = 347 \, \text{Hz} \).

### Final Answer:
The natural frequency of the second tuning fork is \( 347 \, \text{Hz} \).

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