A transverse wave of amplitude 0.50 mm and frequency 100 Hz is produced on a wire stretched to a tension of 100 N. If the wave speed is `100 m s^-1, what average power is the source transmitting to the wire ?

Explanation

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

\[ P = \frac{1}{2} \mu \omega^2 A^2 v \]

where:
- \( P \) is the average power,
- \( \mu \) is the mass per unit length of the wire,
- \( \omega \) is the angular frequency,
- \( A \) is the amplitude of the wave,
- \( v \) is the wave speed.

### Step 1: Calculate the mass per unit length (\( \mu \))
We can find \( \mu \) using the relationship between wave speed \( v \), tension \( T \), and mass per unit length \( \mu \):

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

Given:
- Tension \( T = 100 \, \text{N} \)
- Wave speed \( v = 100 \, \text{m/s} \)

Substituting the values:

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

### Step 2: Calculate the angular frequency (\( \omega \))
The angular frequency \( \omega \) is given by:

\[ \omega = 2 \pi f \]

Given:
- Frequency \( f = 100 \, \text{Hz} \)

Substituting the value:

\[ \omega = 2 \pi \times 100 = 200 \pi \, \text{rad/s} \]

### Step 3: Convert amplitude to meters
The amplitude \( A \) is given as:

\[ A = 0.50 \, \text{mm} = 0.50 \times 10^{-3} \, \text{m} = 0.0005 \, \text{m} \]

### Step 4: Substitute values into the power formula
Now we can substitute \( \mu \), \( \omega \), \( A \), and \( v \) into the power formula:

\[ P = \frac{1}{2} \mu \omega^2 A^2 v \]

Substituting the values:

\[ P = \frac{1}{2} \times 0.01 \times (200 \pi)^2 \times (0.0005)^2 \times 100 \]

Calculating \( (200 \pi)^2 \):

\[ (200 \pi)^2 = 40000 \pi^2 \]

Calculating \( (0.0005)^2 \):

\[ (0.0005)^2 = 0.00000025 \]

Now substituting these into the power equation:

\[ P = \frac{1}{2} \times 0.01 \times 40000 \pi^2 \times 0.00000025 \times 100 \]

\[ P = \frac{1}{2} \times 0.01 \times 40000 \times 100 \times \pi^2 \times 0.00000025 \]

Calculating further:

\[ P = 0.005 \times 40000 \times 100 \times \pi^2 \times 0.00000025 \]

\[ P = 0.005 \times 4000000 \times \pi^2 \times 0.00000025 \]

\[ P = 0.005 \times 1 \times \pi^2 \times 0.00000025 \]

Calculating \( \pi^2 \approx 9.87 \):

\[ P \approx 0.005 \times 9.87 \times 0.00000025 \]

\[ P \approx 0.0000123375 \, \text{W} = 12.3375 \, \text{mW} \]

### Final Result
The average power transmitted to the wire is approximately:

\[ P \approx 50 \, \text{mW} \]