A tuning fork of frequency 512 Hz makes 4 beats//s with the vibrating string of a piano. The beat frequency decreases to 2 beats//s when the tension in the piano string is slightly increased.The frequency of the piano string before increasing the tension was

Options

A. `510 Hz`
B. `514//Hz`
C. `516 Hz`
D. `508 Hz`

Explanation

To solve the problem, we need to determine the frequency of the piano string before the tension was increased. We will use the concept of beat frequency, which is the difference between the frequencies of two vibrating sources.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Frequency of the tuning fork (NF) = 512 Hz
- Initial beat frequency = 4 beats/s
- New beat frequency after increasing tension = 2 beats/s

2. **Understanding Beat Frequency:**
- The beat frequency is the absolute difference between the frequencies of the two sources.
- Let the frequency of the piano string be denoted as (NP).
- The initial beat frequency can be expressed as:
\[ |NF - NP| = 4 \text{ Hz} \]
- This means that the frequency of the piano string could be either:
\[ NP = NF - 4 \quad \text{or} \quad NP = NF + 4 \]

3. **Calculate Possible Frequencies:**
- Using the first case:
\[ NP = 512 - 4 = 508 \text{ Hz} \]
- Using the second case:
\[ NP = 512 + 4 = 516 \text{ Hz} \]

4. **Consider the Effect of Increasing Tension:**
- When the tension in the piano string is increased, the frequency of the string (NP) will increase.
- The new beat frequency is now 2 beats/s, which gives us:
\[ |NF - NP'| = 2 \text{ Hz} \]
- Where NP' is the new frequency of the piano string after increasing tension. Since NP increases, NP' will be greater than NP.

5. **Determine the New Frequency:**
- If we take NP = 508 Hz, then:
\[ NP' = 508 + \Delta \text{ (where } \Delta \text{ is the increase in frequency)} \]
- The new beat frequency equation becomes:
\[ |512 - (508 + \Delta)| = 2 \]
Simplifying this gives:
\[ |4 - \Delta| = 2 \]
- This results in two scenarios:
- \(4 - \Delta = 2 \Rightarrow \Delta = 2\)
- \(4 - \Delta = -2 \Rightarrow \Delta = 6\) (not possible since frequency cannot decrease)

6. **Final Frequency Calculation:**
- Thus, if NP = 508 Hz, then after an increase of 2 Hz, NP' becomes:
\[ NP' = 508 + 2 = 510 \text{ Hz} \]
- The beat frequency with 510 Hz and 512 Hz is:
\[ |512 - 510| = 2 \text{ Hz} \]
- This confirms that NP = 508 Hz is correct.

### Conclusion:
The frequency of the piano string before increasing the tension was **508 Hz**.