A transverse wave is described by the equation
`y=y_(0)sin4pi(upsilont-(x)/(lambda))`.
The maximum particle velocity is equal to four times the wave velocity if

Options

A. `lambda=piy_(0)//4`
B. `lambda=2piy_(0)`
C. `lambda=pi//y_(0)`
D. `lambda=piy_(0)`

Answer

To solve the problem, we need to analyze the given wave equation and derive the necessary relationships between the maximum particle velocity and the wave velocity.

### Step-by-Step Solution:

1. **Identify the Wave Equation:**
The given wave equation is:
\[ y = y_0 \sin\left(4\pi\left(\nu t - \frac{x}{\lambda}\right)\right) \]
Here, \(y_0\) is the amplitude, \(\nu\) is the frequency, and \(\lambda\) is the wavelength.

2. **Determine Angular Frequency (\(\omega\)) and Wave Number (\(k\)):**
From the wave equation, we can identify:
- Angular frequency \(\omega = 4\pi\nu\)
- Wave number \(k = \frac{2\pi}{\lambda}\)

3. **Calculate Wave Velocity (\(v\)):**
The wave velocity \(v\) can be calculated using the formula:
\[ v = \frac{\omega}{k} \]
Substituting the values of \(\omega\) and \(k\):
\[ v = \frac{4\pi\nu}{\frac{2\pi}{\lambda}} = 2\nu\lambda \]

4. **Find the Particle Velocity:**
The particle velocity \(v_p\) is the rate of change of displacement with respect to time. We find it by differentiating \(y\) with respect to \(t\):
\[ v_p = \frac{\partial y}{\partial t} = y_0 \cdot \frac{\partial}{\partial t}\left(\sin(4\pi(\nu t - \frac{x}{\lambda}))\right) \]
Using the chain rule:
\[ v_p = y_0 \cdot 4\pi\nu \cos(4\pi(\nu t - \frac{x}{\lambda})) \]

5. **Determine Maximum Particle Velocity (\(V_{max}\)):**
The maximum particle velocity occurs when \(\cos(4\pi(\nu t - \frac{x}{\lambda})) = 1\):
\[ V_{max} = y_0 \cdot 4\pi\nu \]

6. **Set Up the Relationship:**
According to the problem, the maximum particle velocity is equal to four times the wave velocity:
\[ V_{max} = 4v \]
Substituting the expressions we derived:
\[ y_0 \cdot 4\pi\nu = 4(2\nu\lambda) \]

7. **Simplify the Equation:**
Simplifying the equation gives:
\[ y_0 \cdot 4\pi\nu = 8\nu\lambda \]
Dividing both sides by \(4\nu\) (assuming \(\nu \neq 0\)):
\[ y_0 \pi = 2\lambda \]

8. **Final Result:**
Rearranging gives:
\[ \lambda = \frac{y_0 \pi}{2} \]

### Summary:
The condition for the maximum particle velocity to be equal to four times the wave velocity is:
\[ \lambda = \frac{y_0 \pi}{2} \]