When a tuning fork vibrates with `1.0 m` or `1.05 m` long wire of a , `5` beats per second are produced in each case. If the frequency of the tuning fork is `5f` (in `Hz`) find `f`.

Explanation

To solve the problem step by step, we will follow the reasoning laid out in the video transcript:

### Step 1: Understand the Frequency Formula
The frequency of a vibrating string is given by the formula:
\[ f = \frac{V}{2L} \]
where \( V \) is the speed of the wave on the string, and \( L \) is the length of the string.

### Step 2: Calculate Frequencies for Each Length
1. For a wire of length \( L_1 = 1.0 \, m \):
\[ f_1 = \frac{V}{2 \times 1} = \frac{V}{2} \]

2. For a wire of length \( L_2 = 1.05 \, m \):
\[ f_2 = \frac{V}{2 \times 1.05} = \frac{V}{2.1} \]

### Step 3: Set Up the Beat Frequency Equations
The problem states that 5 beats per second are produced in each case. This means:
1. The difference between the frequencies \( f_1 \) and \( f_2 \) is 5 Hz:
\[ |f_1 - f_2| = 5 \]

### Step 4: Express the Beat Frequency
From the first equation:
\[ f_1 - f_2 = 5 \]
Substituting the values of \( f_1 \) and \( f_2 \):
\[ \frac{V}{2} - \frac{V}{2.1} = 5 \]

### Step 5: Solve for \( V \)
To solve the equation, we need a common denominator:
\[ \frac{V \cdot 2.1 - V \cdot 2}{2 \cdot 2.1} = 5 \]
This simplifies to:
\[ \frac{0.1V}{4.2} = 5 \]
Now, multiply both sides by 4.2:
\[ 0.1V = 21 \]
Dividing both sides by 0.1 gives:
\[ V = 210 \, m/s \]

### Step 6: Calculate \( f_1 \) and \( f_2 \)
Now we can calculate \( f_1 \) and \( f_2 \):
1. For \( f_1 \):
\[ f_1 = \frac{210}{2} = 105 \, Hz \]

2. For \( f_2 \):
\[ f_2 = \frac{210}{2.1} = 100 \, Hz \]

### Step 7: Find the Value of \( f \)
From the problem, we know that:
\[ f_1 - f_2 = 5 \, Hz \]
Now, since \( f_2 = 5f \):
\[ 100 = 5f \]
Thus, solving for \( f \):
\[ f = \frac{100}{5} = 20 \, Hz \]

### Final Answer
The value of \( f \) is:
\[ \boxed{20 \, Hz} \]