the fundamental frequency of a sonometer wire of length is `f_(0)`.A bridge is now introduced at a distance of `Deltal`from the centre of the wire `(Deltal lt lt l )`. The number of beats heard if their fundamental mode are

Options

A. `(8f_(0)Delta|)/|`
B. `(f_(0)Delta|)/|`
C. `(2f_(0)Delta|)/|`
D. `(4f_(0)Delta|)/|`

Answer

To solve the problem step by step, we need to analyze the situation involving the sonometer wire and the introduction of a bridge.

### Step 1: Understand the Fundamental Frequency
The fundamental frequency \( f_0 \) of the sonometer wire of length \( l \) is given by the formula:
\[ f_0 = \frac{K}{L} \]
where \( K \) is a constant related to the tension and mass per unit length of the wire.

### Step 2: Effect of the Bridge
When a bridge is introduced at a distance \( \Delta l \) from the center of the wire, it effectively divides the wire into two segments. The lengths of these segments will be:
- Length of the left segment: \( \frac{l}{2} - \Delta l \)
- Length of the right segment: \( \frac{l}{2} + \Delta l \)

### Step 3: Calculate the Frequencies of the Segments
The fundamental frequencies for the two segments can be calculated as follows:
- For the left segment:
\[ f_1 = \frac{K}{\frac{l}{2} - \Delta l} \]
- For the right segment:
\[ f_2 = \frac{K}{\frac{l}{2} + \Delta l} \]

### Step 4: Find the Difference in Frequencies
The number of beats heard is equal to the difference in frequencies \( |f_1 - f_2| \):
\[ |f_1 - f_2| = \left| \frac{K}{\frac{l}{2} - \Delta l} - \frac{K}{\frac{l}{2} + \Delta l} \right| \]

### Step 5: Simplify the Expression
We can factor out \( K \) from the expression:
\[ |f_1 - f_2| = K \left| \frac{1}{\frac{l}{2} - \Delta l} - \frac{1}{\frac{l}{2} + \Delta l} \right| \]
To simplify the right-hand side:
\[ \frac{1}{\frac{l}{2} - \Delta l} - \frac{1}{\frac{l}{2} + \Delta l} = \frac{(\frac{l}{2} + \Delta l) - (\frac{l}{2} - \Delta l)}{(\frac{l}{2} - \Delta l)(\frac{l}{2} + \Delta l)} = \frac{2 \Delta l}{\left(\frac{l}{2}\right)^2 - (\Delta l)^2} \]
For small \( \Delta l \) compared to \( l \), we can approximate:
\[ \left(\frac{l}{2}\right)^2 - (\Delta l)^2 \approx \left(\frac{l}{2}\right)^2 \]
Thus:
\[ |f_1 - f_2| \approx K \cdot \frac{2 \Delta l}{\left(\frac{l}{2}\right)^2} = \frac{8K \Delta l}{l^2} \]

### Step 6: Relate Back to the Fundamental Frequency
Since \( f_0 = \frac{K}{l} \), we can express \( K \) in terms of \( f_0 \):
\[ K = f_0 \cdot l \]
Substituting this back into our expression for the difference in frequencies:
\[ |f_1 - f_2| = \frac{8(f_0 \cdot l) \Delta l}{l^2} = \frac{8f_0 \Delta l}{l} \]

### Conclusion
The number of beats heard is:
\[ \text{Number of beats} = |f_1 - f_2| = \frac{8f_0 \Delta l}{l} \]