A tuning fork of frequency 340 Hz produces 5 beats per second with a sonometer wire. If the tension is slightly increased the number of beats becomes 4. The frequency of sonometer wire is

Options

A. 335 Hz
B. 345 Hz
C. 330 Hz
D. 350 Hz

Answer

To solve the problem, we need to determine the frequency of the sonometer wire based on the information given about the beats produced when a tuning fork is struck.

### Step-by-Step Solution:

1. **Understanding Beats**:
The number of beats per second is the absolute difference between the frequencies of two sources. In this case, the tuning fork and the sonometer wire.

2. **Initial Condition**:
The tuning fork has a frequency \( f_t = 340 \, \text{Hz} \) and produces 5 beats per second with the sonometer wire.
This can be expressed as:
\[ |f_t - f_s| = 5 \]
where \( f_s \) is the frequency of the sonometer wire.

3. **Setting Up the Equation**:
Since the tuning fork frequency is higher than the sonometer wire frequency (as it produces beats), we can write:
\[ 340 - f_s = 5 \]
Rearranging this gives:
\[ f_s = 340 - 5 = 335 \, \text{Hz} \]

4. **Change in Condition**:
When the tension of the wire is slightly increased, the frequency of the sonometer wire increases, and the number of beats decreases to 4.
Now we have:
\[ |f_t - f_s'| = 4 \]
where \( f_s' \) is the new frequency of the sonometer wire after the tension is increased.

5. **Setting Up the New Equation**:
Again, since the tuning fork frequency is still higher, we can write:
\[ 340 - f_s' = 4 \]
Rearranging gives:
\[ f_s' = 340 - 4 = 336 \, \text{Hz} \]

6. **Conclusion**:
The frequency of the sonometer wire before the tension was increased was:
\[ f_s = 335 \, \text{Hz} \]

### Final Answer:
The frequency of the sonometer wire is \( 335 \, \text{Hz} \).