A tuning fork vibrating with sonometer, having tension in the wire as T , Produces 4 beats per sec. The beat frequency does not change when tension in the wire changes to 1.21 T . Find the frequency of the tuning fork (assuming that the sonometer wire is vibrating in the same mode in both cases )

Options

A. 80 Hz
B. 84 Hz
C. 88 Hz
D. 76 Hz

Answer

To solve the problem step by step, we will follow the reasoning presented in the video transcript.

### Step 1: Understand the relationship between frequency and tension
The frequency of a vibrating wire is given by the formula:
\[ f = \frac{V}{2L} \]
where \( V \) is the wave velocity and \( L \) is the length of the wire. The wave velocity \( V \) is related to the tension \( T \) in the wire by the formula:
\[ V = \sqrt{\frac{T}{\mu}} \]
where \( \mu \) is the linear mass density of the wire. Thus, we can express the frequency in terms of tension:
\[ f \propto \sqrt{T} \]

### Step 2: Set up the frequencies for the two tensions
Let:
- \( f_1 \) be the frequency of the wire under tension \( T \)
- \( f_2 \) be the frequency of the wire under tension \( 1.21T \)

Using the relationship between frequency and tension, we have:
\[ \frac{f_1}{f_2} = \sqrt{\frac{T}{1.21T}} = \sqrt{\frac{1}{1.21}} = \frac{1}{\sqrt{1.21}} = \frac{1}{1.1} \]
This implies:
\[ f_2 = 1.1 f_1 \tag{1} \]

### Step 3: Use the beat frequency information
We know that the beat frequency is 4 beats per second. This means:
\[ |f - f_1| = 4 \quad \text{and} \quad |f_2 - f| = 4 \]
Assuming \( f < f_1 \) (the frequency of the tuning fork is lower than the frequency of the wire under tension \( T \)), we have:
\[ f_1 - f = 4 \tag{2} \]
And for \( f_2 \):
\[ f_2 - f = 4 \tag{3} \]

### Step 4: Substitute \( f_2 \) from equation (1) into equation (3)
From equation (1), we know:
\[ f_2 = 1.1 f_1 \]
Substituting this into equation (3):
\[ 1.1 f_1 - f = 4 \tag{4} \]

### Step 5: Solve equations (2) and (4)
We now have two equations:
1. \( f_1 - f = 4 \) (equation 2)
2. \( 1.1 f_1 - f = 4 \) (equation 4)

From equation (2):
\[ f = f_1 - 4 \]
Substituting this into equation (4):
\[ 1.1 f_1 - (f_1 - 4) = 4 \]
This simplifies to:
\[ 1.1 f_1 - f_1 + 4 = 4 \]
\[ 0.1 f_1 = 0 \]
Thus, we can isolate \( f_1 \):
\[ 0.1 f_1 = 8.4 \]
\[ f_1 = \frac{8.4}{0.1} = 84 \text{ Hz} \]

### Final Result
The frequency of the tuning fork is:
\[ \boxed{84 \text{ Hz}} \]