A tuning fork of frequency 400 Hz, produces 10 beats/ sec, when sounded with a vibrating sonometer string . What must have been the frequency of the string if a slight increase in tension in the string , produces fewer beats per second than before ?

Options

A. 390 Hz
B. 410 Hz
C. 415 Hz
D. 420 Hz

Explanation

To solve the problem, we need to determine the frequency of the string (ns) given that it produces 10 beats per second with a tuning fork of frequency 400 Hz. Additionally, we need to consider the effect of increasing the tension in the string on the beat frequency.

### Step-by-Step Solution:

1. **Understanding Beats**:
The beat frequency is given by the absolute difference between the frequencies of the two sources. Here, it is given that:
\[ |nt - ns| = 10 \text{ beats/second} \]
where \( nt \) is the frequency of the tuning fork (400 Hz) and \( ns \) is the frequency of the string.

2. **Setting Up the Equation**:
Since the tuning fork frequency is higher than the string frequency (as we will see), we can write:
\[ nt - ns = 10 \]
Substituting \( nt = 400 \) Hz, we have:
\[ 400 - ns = 10 \]

3. **Solving for ns**:
Rearranging the equation gives:
\[ ns = 400 - 10 = 390 \text{ Hz} \]

4. **Effect of Increasing Tension**:
When the tension in the string is increased, the frequency of the string (ns) will increase. According to the problem, this slight increase in tension leads to fewer beats per second. This means that after increasing the tension, the frequency of the string must become closer to the frequency of the tuning fork, which is 400 Hz.

5. **Conclusion**:
Since increasing the tension decreases the beat frequency, it implies that the frequency of the string must be less than 400 Hz initially and must be 390 Hz as calculated. Thus, the frequency of the string is:
\[ ns = 390 \text{ Hz} \]

### Final Answer:
The frequency of the string must have been **390 Hz**.