A tunning fork is frequency 512Hz is vibrated with a sonometer wire and 6 beats per second as heard The beat frequency reduces if the tension in the string of slightly increased. The original frequency of vibration of the string is

Options

A. 506Hz
B. 512Hz
C. 518Hz
D. 524Hz

Answer

To solve the problem step by step, we will follow these logical steps:

### Step 1: Understand the given information
We know that:
- The frequency of the tuning fork (F1) is 512 Hz.
- The beat frequency (FB) is 6 beats per second.
- The beat frequency decreases when the tension in the string is slightly increased.

### Step 2: Use the formula for beat frequency
The beat frequency is given by the formula:
\[ FB = |F1 - F2| \]
where F1 is the frequency of the tuning fork and F2 is the frequency of the string.

### Step 3: Set up the equation
From the information provided:
\[ FB = 6 \text{ Hz} \]
So we can write:
\[ 6 = |512 - F2| \]

### Step 4: Solve for F2
This absolute value equation can be split into two cases:
1. \( 512 - F2 = 6 \)
2. \( F2 - 512 = 6 \)

#### Case 1:
\[ 512 - F2 = 6 \]
\[ F2 = 512 - 6 \]
\[ F2 = 506 \text{ Hz} \]

#### Case 2:
\[ F2 - 512 = 6 \]
\[ F2 = 512 + 6 \]
\[ F2 = 518 \text{ Hz} \]

### Step 5: Determine the correct frequency
Since the problem states that the beat frequency decreases when the tension in the string is increased, this implies that the original frequency of the string (F2) must be lower than the frequency of the tuning fork (F1). Therefore, we discard the second case (518 Hz) as it does not fit the condition.

### Conclusion
The original frequency of vibration of the string (F2) is:
\[ F2 = 506 \text{ Hz} \]

### Final Answer
The original frequency of vibration of the string is **506 Hz**.

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