To solve the problem, we will follow these steps:
### Step 1: Write the equation of the wave
The equation for a simple harmonic plane progressive wave traveling along the positive x-axis is given by:
\[ y = A \sin(\omega t - kx) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( k \) is the wave number.
### Step 2: Identify given values
From the question, we know:
- \( \omega = \pi \, \text{rad/s} \)
### Step 3: Determine the wave number \( k \)
We need to find \( k \) using the relationship:
\[ k = \frac{2\pi}{\lambda} \]
To find \( \lambda \), we use the information provided about the particles at \( x = 2 \, \text{m} \) and \( x = 3 \, \text{m} \) at \( t = 1 \, \text{s} \).
### Step 4: Calculate the phase difference
At \( t = 1 \, \text{s} \):
- For \( x = 2 \, \text{m} \):
\[ y(2) = A \sin(\pi \cdot 1 - k \cdot 2) = A \sin(\pi - 2k) \]
- For \( x = 3 \, \text{m} \):
\[ y(3) = A \sin(\pi \cdot 1 - k \cdot 3) = A \sin(\pi - 3k) \]
Since the particle at \( x = 2 \, \text{m} \) is moving downward and the one at \( x = 3 \, \text{m} \) is moving upward, the phase difference \( \Delta \phi \) must be \( \pi \) (indicating they are in opposite directions).
### Step 5: Calculate the path difference
The path difference \( \Delta x \) is:
\[ \Delta x = 3 \, \text{m} - 2 \, \text{m} = 1 \, \text{m} \]
Using the relationship between phase difference and path difference:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
Setting \( \Delta \phi = \pi \):
\[ \pi = \frac{2\pi}{\lambda} \cdot 1 \]
This gives:
\[ \lambda = 2 \, \text{m} \]
### Step 6: Calculate \( k \)
Now substituting \( \lambda \) into the equation for \( k \):
\[ k = \frac{2\pi}{2} = \pi \]
### Step 7: Rewrite the wave equation
Now we can rewrite the wave equation:
\[ y = A \sin(\pi t - \pi x) \]
### Step 8: Find the time for maximum power transfer
The maximum power transfer occurs when the velocity \( v \) is maximized. The velocity is given by:
\[ v = \frac{dy}{dt} = A \pi \cos(\pi t - \pi x) \]
To maximize \( v \), we need:
\[ \pi t - \pi x = \left(2n + 1\right) \frac{\pi}{2} \]
for \( n \in \mathbb{Z} \).
### Step 9: Substitute \( x = \frac{11}{4} \, \text{m} \)
Substituting \( x = \frac{11}{4} \):
\[ \pi t - \pi \cdot \frac{11}{4} = \left(2n + 1\right) \frac{\pi}{2} \]
Dividing through by \( \pi \):
\[ t - \frac{11}{4} = \left(2n + 1\right) \frac{1}{2} \]
This simplifies to:
\[ t = \frac{11}{4} + \frac{2n + 1}{2} \]
### Step 10: Solve for \( t \)
For \( n = 0 \):
\[ t = \frac{11}{4} + \frac{1}{2} = \frac{11}{4} + \frac{2}{4} = \frac{13}{4} \, \text{s} \]
For \( n = 1 \):
\[ t = \frac{11}{4} + 1 = \frac{11}{4} + \frac{4}{4} = \frac{15}{4} \, \text{s} \]
Thus, the time when maximum power transfer will take place at \( x = \frac{11}{4} \, \text{m} \) is:
\[ t = \frac{13}{4} \, \text{s} \text{ or } \frac{15}{4} \, \text{s} \]
