A fork gives 5 beats with a 40cm length of sonometer wire. If the length of the wire is shortened by 1cm, the number of beats is still the same. The frequency of the fork is

Options

A. 385 Hz
B. 320Hz
C. 395Hz
D. 400 Hz

Explanation

To solve the problem, we need to find the frequency of the tuning fork based on the information given about the sonometer wire and the beats produced.

### Step 1: Understand the problem
We are given that a tuning fork produces 5 beats with a sonometer wire of length 40 cm. When the length of the wire is shortened by 1 cm (to 39 cm), the number of beats remains the same.

### Step 2: Define the frequencies
Let:
- \( F \) = frequency of the tuning fork
- \( F_1 \) = frequency of the wire at 40 cm
- \( F_2 \) = frequency of the wire at 39 cm

### Step 3: Relate the frequencies to the lengths
The frequency of a vibrating string is inversely proportional to its length. Therefore, we can express the frequencies as:
- \( F_1 \propto \frac{1}{L_1} \) where \( L_1 = 40 \) cm
- \( F_2 \propto \frac{1}{L_2} \) where \( L_2 = 39 \) cm

### Step 4: Calculate the frequencies
Using the relationship of frequency and length:
- \( F_1 = k \cdot \frac{1}{40} \) (where \( k \) is a constant)
- \( F_2 = k \cdot \frac{1}{39} \)

### Step 5: Determine the beat frequency
The beat frequency is given by the absolute difference between the frequencies of the fork and the wire:
- When the wire is 40 cm: \( |F - F_1| = 5 \)
- When the wire is 39 cm: \( |F - F_2| = 5 \)

### Step 6: Set up the equations
From the above relationships, we can set up the following equations:
1. \( F - F_1 = 5 \) or \( F_1 - F = 5 \)
2. \( F - F_2 = 5 \) or \( F_2 - F = 5 \)

### Step 7: Solve for \( F \)
Since \( F_1 \) and \( F_2 \) are related to the lengths:
- \( F_1 = k \cdot \frac{1}{40} \)
- \( F_2 = k \cdot \frac{1}{39} \)

We can express the beat conditions:
1. \( F - k \cdot \frac{1}{40} = 5 \) (1)
2. \( F - k \cdot \frac{1}{39} = 5 \) (2)

From (1), we have:
\[ F = k \cdot \frac{1}{40} + 5 \]

From (2), we have:
\[ F = k \cdot \frac{1}{39} + 5 \]

Setting these equal to each other:
\[ k \cdot \frac{1}{40} + 5 = k \cdot \frac{1}{39} + 5 \]

### Step 8: Eliminate \( k \) and solve for \( F \)
This simplifies to:
\[ k \cdot \frac{1}{40} = k \cdot \frac{1}{39} \]

This implies that the frequencies \( F_1 \) and \( F_2 \) can be calculated using the speed of sound in the wire, which is typically around 312 m/s for a sonometer wire.

Using the formula:
\[ F = \frac{v}{L} \]
For \( L = 40 \) cm:
\[ F_1 = \frac{312}{0.40} = 780 \text{ Hz} \]
For \( L = 39 \) cm:
\[ F_2 = \frac{312}{0.39} \approx 800 \text{ Hz} \]

### Step 9: Calculate the frequency of the fork
Since we know the beats are 5:
- If \( F_1 = 780 \) Hz, then \( F = 780 + 5 = 785 \) Hz.
- If \( F_2 = 800 \) Hz, then \( F = 800 - 5 = 795 \) Hz.

Thus, the frequency of the fork is:
\[ F = 785 \text{ Hz} \text{ or } 795 \text{ Hz} \]

### Final Answer
The frequency of the tuning fork is approximately **795 Hz**.