A sinusoidal transverse wave travel on a string. The string has length `8.00m` and mass `6.00 g`. The wave speed is `30.0 m//s` and the wavelength is `0.200 m`. (a) If the wave is to have an average power of `50.0 W`, what must be the amplitude of the wave? (b) For the same string, if the ampitude and wavelength are the same as in part (a) what is the average power for the wave if the tension is increased such that the wave speed is doubled?

Explanation

To solve the problem step by step, we will break it down into parts (a) and (b) as given in the question.

### Part (a): Finding the Amplitude

1. **Given Data**:
- Length of the string, \( L = 8.00 \, \text{m} \)
- Mass of the string, \( m = 6.00 \, \text{g} = 0.006 \, \text{kg} \)
- Wave speed, \( v = 30.0 \, \text{m/s} \)
- Wavelength, \( \lambda = 0.200 \, \text{m} \)
- Average power, \( P = 50.0 \, \text{W} \)

2. **Calculate the Linear Mass Density (\( \mu \))**:
\[ \mu = \frac{m}{L} = \frac{0.006 \, \text{kg}}{8.00 \, \text{m}} = 0.00075 \, \text{kg/m} \]

3. **Calculate the Wave Number (\( k \))**:
\[ k = \frac{2\pi}{\lambda} = \frac{2\pi}{0.200 \, \text{m}} = 10\pi \, \text{rad/m} \]

4. **Calculate the Angular Frequency (\( \omega \))**:
Using the relationship \( v = \frac{\omega}{k} \):
\[ \omega = v \cdot k = 30.0 \, \text{m/s} \cdot 10\pi \, \text{rad/m} = 300\pi \, \text{rad/s} \]

5. **Power Formula**:
The average power for a sinusoidal wave on a string is given by:
\[ P = \frac{1}{2} \mu \omega^2 A^2 \]
Rearranging for amplitude \( A \):
\[ A^2 = \frac{2P}{\mu \omega^2} \]

6. **Substitute the Values**:
\[ A^2 = \frac{2 \cdot 50.0 \, \text{W}}{0.00075 \, \text{kg/m} \cdot (300\pi \, \text{rad/s})^2} \]
\[ A^2 = \frac{100.0}{0.00075 \cdot 90000\pi^2} \]
\[ A^2 = \frac{100.0}{0.00075 \cdot 282743.34} \approx \frac{100.0}{212.0575} \approx 0.471 \]
\[ A \approx \sqrt{0.471} \approx 0.686 \, \text{m} \approx 68.6 \, \text{cm} \]

### Part (b): Average Power with Increased Wave Speed

1. **New Wave Speed**:
The wave speed is doubled:
\[ v' = 2v = 2 \cdot 30.0 \, \text{m/s} = 60.0 \, \text{m/s} \]

2. **New Angular Frequency**:
Using the new wave speed:
\[ \omega' = v' \cdot k = 60.0 \, \text{m/s} \cdot 10\pi \, \text{rad/m} = 600\pi \, \text{rad/s} \]

3. **New Power Formula**:
The average power can be recalculated using the same amplitude \( A \) and the new angular frequency:
\[ P' = \frac{1}{2} \mu \omega'^2 A^2 \]
Since \( \omega' = 2\omega \):
\[ P' = \frac{1}{2} \mu (2\omega)^2 A^2 = 4 \cdot \frac{1}{2} \mu \omega^2 A^2 = 4P \]
\[ P' = 4 \cdot 50.0 \, \text{W} = 200.0 \, \text{W} \]

### Final Answers:
(a) The amplitude of the wave is approximately \( 68.6 \, \text{cm} \).

(b) The average power for the wave with doubled wave speed is \( 200.0 \, \text{W} \).