A `200 Hz` wave with amplitude `1 mm` travels on a long string of linear mass density `6 g//m` keep under a tension of `60 N`.
(a) Find the average power transmitted across a given point on the string.
(b) Find the total energy associated with the wave in a `2.0 m` long portion of the string.

Explanation

To solve the problem step by step, we will break it down into two parts: (a) calculating the average power transmitted across a point on the string, and (b) finding the total energy associated with a 2.0 m long portion of the string.

### Part (a): Average Power Transmitted

1. **Identify the given values:**
- Frequency (f) = 200 Hz
- Amplitude (A) = 1 mm = \(1 \times 10^{-3}\) m
- Linear mass density (\(\mu\)) = 6 g/m = \(6 \times 10^{-3}\) kg/m
- Tension (T) = 60 N

2. **Calculate the angular frequency (\(\omega\)):**
\[ \omega = 2\pi f = 2\pi \times 200 = 400\pi \, \text{rad/s} \]

3. **Calculate the wave speed (v):**
The wave speed on a string is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
Substituting the values:
\[ v = \sqrt{\frac{60}{6 \times 10^{-3}}} = \sqrt{10000} = 100 \, \text{m/s} \]

4. **Use the formula for average power (P):**
The average power transmitted by a wave on a string is given by:
\[ P = \frac{1}{2} \mu \omega^2 A^2 v \]
Substituting the known values:
\[ P = \frac{1}{2} \times (6 \times 10^{-3}) \times (400\pi)^2 \times (1 \times 10^{-3})^2 \times 100 \]

5. **Calculate \(\omega^2\) and \(A^2\):**
\[ \omega^2 = (400\pi)^2 = 160000\pi^2 \]
\[ A^2 = (1 \times 10^{-3})^2 = 1 \times 10^{-6} \]

6. **Substituting back into the power formula:**
\[ P = \frac{1}{2} \times (6 \times 10^{-3}) \times (160000\pi^2) \times (1 \times 10^{-6}) \times 100 \]
\[ P = 3 \times 10^{-3} \times 160000\pi^2 \times 10^{-4} \]
\[ P = 0.48 \, \text{W} \]

### Part (b): Total Energy Associated with the Wave

1. **Use the formula for energy density (u):**
The energy density of the wave is given by:
\[ u = \frac{1}{2} \mu \omega^2 A^2 \]

2. **Calculate the energy density:**
\[ u = \frac{1}{2} \times (6 \times 10^{-3}) \times (160000\pi^2) \times (1 \times 10^{-6}) \]
\[ u = 3 \times 10^{-3} \times 160000\pi^2 \times 10^{-6} \]

3. **Calculate the mass of the 2 m portion of the string:**
\[ \text{Mass} = \mu \times \text{length} = (6 \times 10^{-3}) \times 2 = 12 \times 10^{-3} \, \text{kg} \]

4. **Calculate the total energy (E) in the 2 m portion:**
\[ E = u \times \text{length} = u \times 2 \]
\[ E = \left(\frac{1}{2} \times (6 \times 10^{-3}) \times (160000\pi^2) \times (1 \times 10^{-6})\right) \times 2 \]
\[ E = 12 \times 10^{-3} \times 80\pi^2 \times 10^{-6} \]
\[ E \approx 9.6 \, \text{mJ} \]

### Final Answers:
- (a) Average Power \(P \approx 0.48 \, \text{W}\)
- (b) Total Energy \(E \approx 9.6 \, \text{mJ}\)