A travelling wave is described by the equation `y = y_(0) sin2 pi ((ft - (x)/lambda))`. The maximum particle velocity is equal to four times the wave velocity if

Options

A. `lamda = pi (y_0)/(4)`
B. `lamda = pi (y_0)/(2)`
C. `lamda = pi y_0`
D. `lamda = 2 pi y_0`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the wave equation and parameters
The given wave equation is:
\[ y = y_0 \sin\left(2\pi\left(ft - \frac{x}{\lambda}\right)\right) \]

From this equation, we can identify:
- \( y_0 \): Amplitude of the wave
- \( f \): Frequency of the wave
- \( \lambda \): Wavelength of the wave

### Step 2: Calculate the maximum particle velocity
The particle velocity \( v \) can be derived from the wave equation by differentiating \( y \) with respect to time \( t \):
\[ v = \frac{dy}{dt} = 2\pi f y_0 \cos\left(2\pi\left(ft - \frac{x}{\lambda}\right)\right) \]

The maximum particle velocity \( V_{\text{max}} \) occurs when \( \cos\left(2\pi\left(ft - \frac{x}{\lambda}\right)\right) = 1 \):
\[ V_{\text{max}} = 2\pi f y_0 \]

### Step 3: Calculate the wave velocity
The wave velocity \( V \) is given by the formula:
\[ V = f \lambda \]

### Step 4: Set up the equation based on the problem statement
According to the problem, the maximum particle velocity is equal to four times the wave velocity:
\[ V_{\text{max}} = 4V \]

Substituting the expressions we found:
\[ 2\pi f y_0 = 4(f \lambda) \]

### Step 5: Simplify the equation
We can simplify the equation by canceling \( f \) from both sides (assuming \( f \neq 0 \)):
\[ 2\pi y_0 = 4\lambda \]

### Step 6: Solve for \( \lambda \)
Rearranging the equation gives:
\[ \lambda = \frac{2\pi y_0}{4} = \frac{\pi y_0}{2} \]

### Conclusion
Thus, we find that:
\[ \lambda = \frac{\pi y_0}{2} \]

### Final Answer
The maximum particle velocity is equal to four times the wave velocity if:
\[ \lambda = \frac{\pi y_0}{2} \]