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 step by step, we need to analyze the relationship between tension, frequency, and the change in tension that results in a specific change in frequency. Here’s how we can approach the problem:

### Step 1: Understand the given information
We have two identical sonometer wires with:
- Fundamental frequency (F) = 500 Hz
- Beat frequency = 5 beats/s

### Step 2: Relate the beat frequency to the change in frequency
The beat frequency is the absolute difference between the frequencies of the two wires. If we denote the frequency of the first wire as F1 and the frequency of the second wire as F2, we can express this as:
\[ |F1 - F2| = 5 \text{ Hz} \]

### Step 3: Calculate the frequency of the second wire
Assuming the first wire has a frequency of 500 Hz (F1 = 500 Hz), we can express the frequency of the second wire (F2) as:
\[ F2 = F1 + 5 \text{ Hz} \text{ or } F2 = F1 - 5 \text{ Hz} \]
Thus:
\[ F2 = 505 \text{ Hz} \text{ or } F2 = 495 \text{ Hz} \]

### Step 4: Use the relationship between frequency and tension
The frequency of a wire under tension is given by the formula:
\[ F = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
Where:
- \( F \) = frequency
- \( L \) = length of the wire (constant)
- \( T \) = tension
- \( \mu \) = mass per unit length (constant)

### Step 5: Express the change in frequency in terms of change in tension
Let’s denote the original tension as \( T_0 \) for the first wire, and the new tension for the second wire as \( T_1 \). The frequencies can be expressed as:
\[ F1 = \frac{1}{2L} \sqrt{\frac{T_0}{\mu}} \]
\[ F2 = \frac{1}{2L} \sqrt{\frac{T_1}{\mu}} \]

### Step 6: Set up the equation for the change in frequency
Using the frequencies we found:
\[ 500 = \frac{1}{2L} \sqrt{\frac{T_0}{\mu}} \]
\[ 505 = \frac{1}{2L} \sqrt{\frac{T_1}{\mu}} \]

### Step 7: Relate the tensions
From the frequency equations, we can express \( T_0 \) and \( T_1 \):
\[ T_0 = (2L \cdot 500)^2 \mu \]
\[ T_1 = (2L \cdot 505)^2 \mu \]

### Step 8: Calculate the change in tension
Now we can find the change in tension:
\[ \Delta T = T_1 - T_0 \]
\[ \Delta T = (2L \cdot 505)^2 \mu - (2L \cdot 500)^2 \mu \]

### Step 9: Factor out common terms
\[ \Delta T = \mu (2L)^2 (505^2 - 500^2) \]

### Step 10: Calculate the percentage change in tension
The percentage change in tension is given by:
\[ \text{Percentage change} = \frac{\Delta T}{T_0} \times 100 \]

### Step 11: Substitute and simplify
Using the values:
\[ T_0 = (2L \cdot 500)^2 \mu \]
We can substitute \( \Delta T \) and simplify to find the percentage change.

After calculating, we find that:
\[ \text{Percentage change} = 2\% \]

### Final Answer
The percentage change in tension of one of the wires that would cause an occurrence of 5 beats/s is **2%**.

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