A tuning fork of known frequency `256 Hz` makes `5` beats per second with the vibrating string of a piano. The beat frequency decreases to `2` beats per second when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was

Options

A. (a)`(256+2)Hz`
B. (b)`(256-2)Hz`
C. (c)`(256-5)Hz`
D. (d)`(256+5)Hz`

Explanation

To solve the problem, we need to determine the frequency of the piano string before the tension was increased. Let's break it down step by step.

### Step 1: Understand the Beat Frequency
The beat frequency is the difference between the frequency of the tuning fork and the frequency of the vibrating string. Given that the tuning fork has a frequency of \( f_t = 256 \, \text{Hz} \) and it makes \( 5 \, \text{beats/second} \) with the piano string, we can express this relationship mathematically.

### Step 2: Calculate Possible Frequencies of the Piano String
The frequency of the piano string \( f_p \) can be either:
1. \( f_p = f_t + 5 \, \text{Hz} = 256 \, \text{Hz} + 5 \, \text{Hz} = 261 \, \text{Hz} \)
2. \( f_p = f_t - 5 \, \text{Hz} = 256 \, \text{Hz} - 5 \, \text{Hz} = 251 \, \text{Hz} \)

So, the possible frequencies of the piano string before increasing the tension are \( 261 \, \text{Hz} \) or \( 251 \, \text{Hz} \).

### Step 3: Analyze the Effect of Increasing Tension
When the tension in the piano string is increased, the frequency of the string increases. The problem states that the beat frequency decreases to \( 2 \, \text{beats/second} \).

If we assume the initial frequency of the piano string was \( 261 \, \text{Hz} \):
- After increasing tension, the frequency would be greater than \( 261 \, \text{Hz} \).
- The beat frequency would then be \( |261 - 256| = 5 \, \text{Hz} \), which contradicts the given information that the beat frequency decreased to \( 2 \, \text{Hz} \).

If we assume the initial frequency of the piano string was \( 251 \, \text{Hz} \):
- After increasing tension, the frequency would be greater than \( 251 \, \text{Hz} \).
- The beat frequency would then be \( |256 - 251| = 5 \, \text{Hz} \) initially, and after increasing the tension, the frequency could be \( 253 \, \text{Hz} \) (for example), resulting in a beat frequency of \( |256 - 253| = 3 \, \text{Hz} \), which is still not \( 2 \, \text{Hz} \).

However, if we assume the frequency increased to \( 254 \, \text{Hz} \), then:
- The beat frequency would be \( |256 - 254| = 2 \, \text{Hz} \), which matches the given condition.

### Conclusion
Thus, the frequency of the piano string before increasing the tension was \( 251 \, \text{Hz} \).

### Final Answer
The frequency of the piano string before increasing the tension was \( 251 \, \text{Hz} \).

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