A transverse wave of amplitude 5 mm and frequency 10Hz is produced on a wire stretched to a tension of 100N. If the wave speed is `100m//s`, what average power is the source transmitting to the wire? `(pi^2 = 10)`

Options

A. 5mW
B. 50 mW
C. 100 mW
D. 150 mW

Explanation

To find the average power transmitted to the wire by the source, we can use the formula for average power in a transverse wave:

\[ P = 2 \pi^2 F^2 A^2 \mu V \]

where:
- \( P \) = average power
- \( F \) = frequency of the wave
- \( A \) = amplitude of the wave
- \( \mu \) = mass per unit length of the wire
- \( V \) = wave speed

### Step 1: Convert the given values to appropriate units
- Amplitude \( A = 5 \, \text{mm} = 5 \times 10^{-3} \, \text{m} \)
- Frequency \( F = 10 \, \text{Hz} \)
- Tension \( T = 100 \, \text{N} \)
- Wave speed \( V = 100 \, \text{m/s} \)

### Step 2: Calculate the mass per unit length \( \mu \)
The mass per unit length \( \mu \) can be calculated using the relationship between wave speed, tension, and mass per unit length:

\[ V = \sqrt{\frac{T}{\mu}} \implies V^2 = \frac{T}{\mu} \implies \mu = \frac{T}{V^2} \]

Substituting the known values:

\[ \mu = \frac{100 \, \text{N}}{(100 \, \text{m/s})^2} = \frac{100}{10000} = 0.01 \, \text{kg/m} \]

### Step 3: Substitute the values into the power formula
Now, substituting \( A \), \( F \), \( \mu \), and \( V \) into the power formula:

\[ P = 2 \pi^2 F^2 A^2 \mu V \]

Using \( \pi^2 = 10 \):

\[ P = 2 \times 10 \times (10)^2 \times (5 \times 10^{-3})^2 \times (0.01) \times (100) \]

Calculating each term step by step:

1. \( F^2 = 10^2 = 100 \)
2. \( A^2 = (5 \times 10^{-3})^2 = 25 \times 10^{-6} \)
3. Substitute these into the equation:

\[ P = 2 \times 10 \times 100 \times 25 \times 10^{-6} \times 0.01 \times 100 \]

### Step 4: Simplify the expression
Calculating the constants:

\[ P = 2 \times 10 \times 100 \times 25 \times 10^{-6} \times 0.01 \times 100 \]

Calculating step by step:

- \( 2 \times 10 = 20 \)
- \( 20 \times 100 = 2000 \)
- \( 2000 \times 25 = 50000 \)
- \( 50000 \times 10^{-6} = 0.05 \)
- \( 0.05 \times 100 = 5 \)

Thus, the average power \( P = 5 \times 10^{-2} \, \text{W} \) or \( 50 \, \text{mW} \).

### Final Answer
The average power transmitted to the wire is \( 50 \, \text{mW} \).

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