A string under a tension of 129.6 N produces 10 beats per sec when it is vibrated along with a tuning fork. When the tension in the string is increased to 160 N, it sounds in unison with the same tuning fork, Calculate the fundamental frequency of the tuning fork.

Options

A. 88 Hz
B. 100 Hz
C. 110 Hz
D. 121 Hz

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between tension and frequency
The frequency of a vibrating string is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( f \) = frequency of the string
- \( L \) = length of the string (constant in this case)
- \( T \) = tension in the string
- \( \mu \) = mass per unit length of the string (constant in this case)

### Step 2: Set up the equations for the two tensions
Let:
- \( T_1 = 129.6 \, \text{N} \)
- \( T_2 = 160 \, \text{N} \)
- \( f_1 \) = frequency of the string at tension \( T_1 \)
- \( f_2 \) = frequency of the string at tension \( T_2 \)
- \( f_0 \) = fundamental frequency of the tuning fork
- \( b = 10 \, \text{beats/sec} \)

From the problem, we know:
1. When the tension is \( T_1 \), the string produces beats with the tuning fork:
\[ f_1 = f_0 - b \]
Thus,
\[ f_1 = f_0 - 10 \] (Equation 1)

2. When the tension is \( T_2 \), the string sounds in unison with the tuning fork:
\[ f_2 = f_0 \] (Equation 2)

### Step 3: Express the frequencies in terms of tension
Using the frequency formula, we can express \( f_1 \) and \( f_2 \):
\[ f_1 = \frac{1}{2L} \sqrt{\frac{T_1}{\mu}} \]
\[ f_2 = \frac{1}{2L} \sqrt{\frac{T_2}{\mu}} \]

### Step 4: Set up the ratio of the frequencies
Now we can set up the ratio of \( f_1 \) and \( f_2 \):
\[ \frac{f_1}{f_2} = \frac{\sqrt{T_1}}{\sqrt{T_2}} \]
Substituting \( f_1 \) and \( f_2 \):
\[ \frac{f_0 - 10}{f_0} = \frac{\sqrt{129.6}}{\sqrt{160}} \]

### Step 5: Solve for \( f_0 \)
Cross-multiplying gives:
\[ (f_0 - 10) \sqrt{160} = f_0 \sqrt{129.6} \]
Expanding this:
\[ f_0 \sqrt{160} - 10 \sqrt{160} = f_0 \sqrt{129.6} \]
Rearranging terms:
\[ f_0 \sqrt{160} - f_0 \sqrt{129.6} = 10 \sqrt{160} \]
Factoring out \( f_0 \):
\[ f_0 \left( \sqrt{160} - \sqrt{129.6} \right) = 10 \sqrt{160} \]
Thus, we can solve for \( f_0 \):
\[ f_0 = \frac{10 \sqrt{160}}{\sqrt{160} - \sqrt{129.6}} \]

### Step 6: Calculate \( f_0 \)
Calculating the square roots:
- \( \sqrt{160} \approx 12.65 \)
- \( \sqrt{129.6} \approx 11.39 \)

Now substituting these values:
\[ f_0 = \frac{10 \times 12.65}{12.65 - 11.39} = \frac{126.5}{1.26} \approx 100 \, \text{Hz} \]

### Conclusion
The fundamental frequency of the tuning fork is \( f_0 = 100 \, \text{Hz} \).