Two identical sonometer wires have a fundamental frequency of `500 Hz` when kept under the same tension . The percentage change in tension of one of the wires that would cause an occurrence of `5 beats//s` , when both wires vibrate together is

Options

A. `0.5 %`
B. `1 %`
C. `2 %`
D. `4 %`

Explanation

To solve the problem, we need to determine the percentage change in tension of one of the wires that would cause a beat frequency of 5 beats per second when both wires vibrate together. We start with the fundamental frequency formula for a wire:

### Step 1: Understand the relationship between frequency and tension
The fundamental frequency \( f \) of a wire fixed at both ends is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( L \) is the length of the wire,
- \( T \) is the tension in the wire,
- \( \mu \) is the linear mass density of the wire.

### Step 2: Define the original frequency and tension
Given that both wires have the same fundamental frequency of \( 500 \, \text{Hz} \) under the same tension, we can denote the original tension as \( T_0 \).

### Step 3: Introduce the change in tension
Let the tension in one of the wires change to \( T_1 \). The new frequency \( f_1 \) of the wire with changed tension can be expressed as:
\[ f_1 = \frac{1}{2L} \sqrt{\frac{T_1}{\mu}} \]

### Step 4: Calculate the beat frequency
The beat frequency \( \Delta f \) is given by the absolute difference between the two frequencies:
\[ \Delta f = |f_1 - f_2| \]
where \( f_2 = 500 \, \text{Hz} \) (the frequency of the unchanged wire). We know that \( \Delta f = 5 \, \text{Hz} \).

### Step 5: Relate the change in frequency to the change in tension
Using the formula for the change in frequency due to a change in tension, we can apply the concept of relative change:
\[ \frac{\Delta f}{f} = \frac{1}{2} \frac{\Delta T}{T} \]
where \( \Delta T = T_1 - T_0 \).

### Step 6: Substitute known values
Substituting \( \Delta f = 5 \, \text{Hz} \) and \( f = 500 \, \text{Hz} \):
\[ \frac{5}{500} = \frac{1}{2} \frac{\Delta T}{T_0} \]
This simplifies to:
\[ \frac{1}{100} = \frac{1}{2} \frac{\Delta T}{T_0} \]

### Step 7: Solve for the change in tension
Rearranging gives:
\[ \Delta T = \frac{2}{100} T_0 = \frac{T_0}{50} \]

### Step 8: Calculate the percentage change in tension
The percentage change in tension is given by:
\[ \text{Percentage change} = \frac{\Delta T}{T_0} \times 100\% \]
Substituting \( \Delta T = \frac{T_0}{50} \):
\[ \text{Percentage change} = \frac{T_0/50}{T_0} \times 100\% = \frac{1}{50} \times 100\% = 2\% \]

### Conclusion
Thus, the percentage change in tension that would cause an occurrence of 5 beats per second is **2%**.

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