Two tuning forks, one of frequency `340 Hz` and second of some unknown frequency produces `5` beats. When filling is done to second fork, then they produces `2` beats. Find initial frequency of second tuning fork.

Options

A. `342 Hz`
B. `338 Hz`
C. `345 Hz`
D. `335 Hz`

Answer

To solve the problem, we need to find the initial frequency of the second tuning fork (f2) based on the information given about the beats produced when the two forks are played together.

### Step-by-Step Solution:

1. **Identify Known Frequencies:**
- The frequency of the first tuning fork (f1) is given as 340 Hz.
- The unknown frequency of the second tuning fork is denoted as f2.

2. **Understanding Beats:**
- When two tuning forks are played together, the number of beats produced is equal to the absolute difference in their frequencies.
- Initially, the two forks produce 5 beats, which means:
\[ |f1 - f2| = 5 \text{ Hz} \]

3. **Setting Up Equations:**
- From the beat frequency, we can derive two possible equations for f2:
\[ f2 = f1 - 5 \quad \text{(Case 1)} \]
\[ f2 = f1 + 5 \quad \text{(Case 2)} \]
- Substituting f1 = 340 Hz:
- Case 1:
\[ f2 = 340 - 5 = 335 \text{ Hz} \]
- Case 2:
\[ f2 = 340 + 5 = 345 \text{ Hz} \]

4. **Filing Effect on Frequency:**
- When the second tuning fork is filed, its frequency increases. The problem states that after filing, the two forks produce 2 beats:
\[ |f1 - f2'| = 2 \text{ Hz} \]
- Here, f2' is the new frequency of the second tuning fork after filing.

5. **Analyzing the Cases:**
- If f2 = 335 Hz (Case 1):
- After filing, f2' increases, so:
\[ f2' > 335 \text{ Hz} \]
- The new frequency difference becomes:
\[ |340 - f2'| = 2 \text{ Hz} \implies f2' = 340 - 2 = 338 \text{ Hz} \quad \text{or} \quad f2' = 340 + 2 = 342 \text{ Hz} \]
- Since f2' must be greater than 335 Hz, the valid option is:
\[ f2' = 342 \text{ Hz} \]
- The difference in frequencies before and after filing is:
\[ |340 - 335| = 5 \text{ Hz} \quad \text{and} \quad |340 - 342| = 2 \text{ Hz} \]
- This case is consistent with the problem statement.

- If f2 = 345 Hz (Case 2):
- After filing, f2' increases:
\[ f2' > 345 \text{ Hz} \]
- The new frequency difference becomes:
\[ |340 - f2'| = 2 \text{ Hz} \implies f2' = 340 - 2 = 338 \text{ Hz} \quad \text{or} \quad f2' = 340 + 2 = 342 \text{ Hz} \]
- Here, both possible values for f2' are less than f2, which contradicts the filing effect (frequency should increase).

6. **Conclusion:**
- The only consistent case is when f2 = 335 Hz. Therefore, the initial frequency of the second tuning fork is:
\[ \boxed{335 \text{ Hz}} \]