A trasverse wave of amplitude `0.50 mm` and frequency `100Hz` is produced on a wire stretched to a tension of `100 N`. If the wave speed is `100 m//s`. What average power is the source transmitting to the wire?

Explanation

To find the average power transmitted by the source to the wire, we can follow these steps:

### Step 1: Identify the given values
- Amplitude (A) = 0.50 mm = 0.50 × 10^-3 m = 5 × 10^-4 m
- Frequency (f) = 100 Hz
- Tension (T) = 100 N
- Wave speed (V) = 100 m/s

### Step 2: Calculate the mass per unit length (μ)
The mass per unit length (μ) can be calculated using the formula:
\[ V = \sqrt{\frac{T}{\mu}} \]
Rearranging this gives us:
\[ \mu = \frac{T}{V^2} \]
Substituting the known values:
\[ \mu = \frac{100 \, \text{N}}{(100 \, \text{m/s})^2} = \frac{100}{10000} = 0.01 \, \text{kg/m} \]

### Step 3: Use the average power formula
The average power (P) transmitted by the wave can be calculated using the formula:
\[ P = 2 \pi^2 \mu V A^2 f^2 \]
Substituting the values we have:
\[ P = 2 \times \pi^2 \times (0.01) \times (100) \times (5 \times 10^{-4})^2 \times (100)^2 \]

### Step 4: Calculate each component
1. Calculate \(A^2\):
\[ A^2 = (5 \times 10^{-4})^2 = 25 \times 10^{-8} = 2.5 \times 10^{-7} \, \text{m}^2 \]
2. Calculate \(f^2\):
\[ f^2 = (100)^2 = 10000 \]
3. Now substitute these values back into the power formula:
\[ P = 2 \times \pi^2 \times (0.01) \times (100) \times (2.5 \times 10^{-7}) \times (10000) \]

### Step 5: Calculate the numerical value
Calculating \(2 \pi^2\):
\[ 2 \pi^2 \approx 19.7392 \]
Now substituting everything:
\[ P \approx 19.7392 \times 0.01 \times 100 \times 2.5 \times 10^{-7} \times 10000 \]
\[ P \approx 19.7392 \times 0.01 \times 100 \times 2.5 \times 10^{-3} \]
\[ P \approx 19.7392 \times 0.025 = 0.49348 \, \text{W} \]
Converting to milliWatts:
\[ P \approx 49.348 \, \text{mW} \approx 49 \, \text{mW} \]

### Final Answer:
The average power transmitted to the wire is approximately **49 mW**.

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