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 is increased. Here is a step-by-step solution:

### Step 1: Understand the Beat Frequency
The beat frequency is the absolute difference between the frequency of the tuning fork and the frequency of the vibrating string. Given:
- Frequency of tuning fork, \( \nu_1 = 256 \, \text{Hz} \)
- Initial beat frequency = 5 beats per second

### Step 2: Set Up the Equation for Initial Beat Frequency
The beat frequency can be expressed as:
\[ |\nu_1 - \nu_2| = 5 \]
This gives us two possible equations:
1. \( \nu_1 - \nu_2 = 5 \)
2. \( \nu_2 - \nu_1 = 5 \)

From these, we can derive:
1. \( \nu_2 = \nu_1 - 5 \)
2. \( \nu_2 = \nu_1 + 5 \)

Substituting \( \nu_1 = 256 \):
1. \( \nu_2 = 256 - 5 = 251 \, \text{Hz} \)
2. \( \nu_2 = 256 + 5 = 261 \, \text{Hz} \)

### 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 beats per second.

### Step 4: Set Up the Equation for New Beat Frequency
After increasing the tension, the new beat frequency is given as:
\[ |\nu_1 - \nu_2'| = 2 \]
Where \( \nu_2' \) is the new frequency of the piano string after increasing tension.

### Step 5: Determine the Possible Cases
1. If \( \nu_2 = 261 \, \text{Hz} \):
- Increasing tension would increase \( \nu_2 \) further, leading to a beat frequency greater than 5, which contradicts the condition of decreasing beat frequency.

2. If \( \nu_2 = 251 \, \text{Hz} \):
- If we assume \( \nu_2' = 255 \, \text{Hz} \) after increasing tension:
\[ |\nu_1 - \nu_2'| = |256 - 255| = 1 \, \text{(not 2)} \]
- If we assume \( \nu_2' = 254 \, \text{Hz} \):
\[ |\nu_1 - \nu_2'| = |256 - 254| = 2 \, \text{(correct)} \]

### Conclusion
Thus, the frequency of the piano string before increasing the tension was:
\[ \nu_2 = 251 \, \text{Hz} \]

### Final Answer
The frequency of the piano string before increasing the tension was **251 Hz**.

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