To find the average rate at which energy is transported by the wave along the cord, we can use the formula for the power transmitted by a sinusoidal wave. The average power \( P \) transmitted by a wave on a string is given by:
\[
P = \frac{1}{2} \mu \omega^2 A^2 v
\]
Where:
- \( \mu \) is the linear density of the cord (in kg/m),
- \( \omega \) is the angular frequency (in rad/s),
- \( A \) is the amplitude of the wave (in m),
- \( v \) is the wave speed (in m/s).
### Step 1: Convert linear density to kg/m
The linear density given is 2.00 g/m. We need to convert this to kg/m:
\[
\mu = 2.00 \, \text{g/m} = 2.00 \times 10^{-3} \, \text{kg/m}
\]
### Step 2: Calculate the wave speed \( v \)
The wave speed \( v \) can be calculated using the formula:
\[
v = \sqrt{\frac{T}{\mu}}
\]
Where \( T \) is the tension in the cord (1200 N). Plugging in the values:
\[
v = \sqrt{\frac{1200 \, \text{N}}{2.00 \times 10^{-3} \, \text{kg/m}}} = \sqrt{600000} \approx 774.6 \, \text{m/s}
\]
### Step 3: Calculate the average power \( P \)
Now we can substitute the values into the power formula. First, we need the angular frequency \( \omega \) and the amplitude \( A \):
- Given \( \omega = 1200 \, \text{rad/s} \)
- Given \( A = 3.00 \, \text{mm} = 3.00 \times 10^{-3} \, \text{m} \)
Now, substituting these values into the power formula:
\[
P = \frac{1}{2} \mu \omega^2 A^2 v
\]
Calculating \( \omega^2 \) and \( A^2 \):
\[
\omega^2 = (1200)^2 = 1440000 \, \text{rad}^2/\text{s}^2
\]
\[
A^2 = (3.00 \times 10^{-3})^2 = 9.00 \times 10^{-6} \, \text{m}^2
\]
Now substituting all values into the power formula:
\[
P = \frac{1}{2} (2.00 \times 10^{-3}) (1440000) (9.00 \times 10^{-6}) (774.6)
\]
Calculating step by step:
1. Calculate \( \frac{1}{2} \times 2.00 \times 10^{-3} = 1.00 \times 10^{-3} \)
2. Calculate \( 1.00 \times 10^{-3} \times 1440000 = 1440 \)
3. Calculate \( 1440 \times 9.00 \times 10^{-6} = 0.01296 \)
4. Finally, calculate \( 0.01296 \times 774.6 \approx 10.05 \, \text{W} \)
### Final Answer
The average rate at which energy is transported by the wave to the opposite end of the cord is approximately:
\[
P \approx 10.05 \, \text{W}
\]
