Two identical wires have the same fundamental frequency of 400 Hz . when kept under the same tension. If the tension in one wire is increased by 2% the number of beats produced will be

Options

A. (a)4
B. (b)8
C. (c)2
D. 1

Answer

To solve the problem step by step, we can follow these steps:

### Step 1: Understand the relationship between frequency and tension
The fundamental frequency \( f \) of a wire is related to the tension \( T \) in the wire by the formula:
\[ f \propto \sqrt{T} \]
This means that the frequency is directly proportional to the square root of the tension.

### Step 2: Write the formula for the change in frequency
If the tension in one wire is increased by a certain percentage, we can express the new frequency \( f' \) as:
\[ f' = f \sqrt{\frac{T'}{T}} \]
where \( T' \) is the new tension and \( T \) is the original tension.

### Step 3: Calculate the new tension
If the original tension \( T \) is increased by 2%, then:
\[ T' = T + 0.02T = 1.02T \]

### Step 4: Calculate the new frequency
Using the relationship from Step 2:
\[ f' = f \sqrt{\frac{1.02T}{T}} = f \sqrt{1.02} \]
Given that the original frequency \( f = 400 \, \text{Hz} \):
\[ f' = 400 \sqrt{1.02} \]

### Step 5: Approximate \( \sqrt{1.02} \)
Using the approximation \( \sqrt{1 + x} \approx 1 + \frac{x}{2} \) for small \( x \):
\[ \sqrt{1.02} \approx 1 + \frac{0.02}{2} = 1 + 0.01 = 1.01 \]
Thus:
\[ f' \approx 400 \times 1.01 = 404 \, \text{Hz} \]

### Step 6: Calculate the beat frequency
The beat frequency \( \Delta f \) is the difference between the two frequencies:
\[ \Delta f = |f' - f| = |404 - 400| = 4 \, \text{Hz} \]

### Conclusion
The number of beats produced will be \( 4 \, \text{Hz} \).