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

Options

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

Answer

To solve the problem, we need to determine the original frequency of vibration of the sonometer wire based on the information given about the tuning fork and the beat frequency. Here’s a step-by-step solution:

### Step 1: Understand the Concept of Beats
When two sound waves of slightly different frequencies interfere, they produce a phenomenon known as beats. The beat frequency is the absolute difference between the two frequencies.

### Step 2: Identify Given Information
- Frequency of the tuning fork (ν1) = 512 Hz
- Beat frequency (n) = 6 beats per second

### Step 3: Set Up the Beat Frequency Equation
The beat frequency can be expressed as:
\[ |ν1 - ν2| = n \]
Where ν2 is the frequency of the sonometer wire. Therefore, we can write two possible equations:
1. \( ν1 - ν2 = n \)
2. \( ν2 - ν1 = n \)

### Step 4: Substitute Known Values
Using the first equation:
\[ ν1 - ν2 = n \]
Substituting the known values:
\[ 512 - ν2 = 6 \]

### Step 5: Solve for ν2
Rearranging the equation gives:
\[ ν2 = 512 - 6 \]
\[ ν2 = 506 \, \text{Hz} \]

### Step 6: Analyze the Effect of Tension on Frequency
The problem states that increasing the tension in the string reduces the beat frequency. This implies that the frequency of the sonometer wire (ν2) must be less than the frequency of the tuning fork (ν1). Since we calculated ν2 = 506 Hz, which is indeed less than 512 Hz, our solution is consistent with the information provided.

### Conclusion
The original frequency of vibration of the string is:
\[ \text{ν2} = 506 \, \text{Hz} \]

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

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