A transverse wave is described by the equatiion `Y = Y_(0) sin 2pi (ft -x//lambda)`. The maximum particle velocity is equal to four times the wave velocity if
`lambda = pi Y_(0) //4`
`lambda = pi Y_(0)//2`
`lambda = pi Y_(0)`
`lambda = 2pi Y_(0)`

Options

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

Explanation

To solve the problem step by step, we will analyze the given wave equation and derive the necessary quantities to find the correct wavelength (\(\lambda\)) that satisfies the condition of maximum particle velocity being equal to four times the wave velocity.

### Step 1: Understand the wave equation
The transverse wave is described by the equation:
\[ Y = Y_0 \sin\left(2\pi\left(ft - \frac{x}{\lambda}\right)\right) \]
where:
- \(Y_0\) is the amplitude,
- \(f\) is the frequency,
- \(x\) is the position,
- \(\lambda\) is the wavelength.

### Step 2: Calculate wave velocity (\(v\))
The wave velocity \(v\) can be calculated using the coefficients of \(t\) and \(x\) in the wave equation:
\[ v = \frac{\text{coefficient of } t}{\text{coefficient of } x} = \frac{2\pi f}{\frac{2\pi}{\lambda}} = \lambda f \]

### Step 3: Calculate particle velocity (\(v_p\))
The particle velocity is found by differentiating the displacement \(Y\) with respect to time \(t\):
\[ v_p = \frac{\partial Y}{\partial t} = Y_0 \cdot \cos\left(2\pi\left(ft - \frac{x}{\lambda}\right)\right) \cdot \frac{\partial}{\partial t}(2\pi(ft - \frac{x}{\lambda})) = Y_0 \cdot \cos\left(2\pi\left(ft - \frac{x}{\lambda}\right)\right) \cdot 2\pi f \]
Thus, the particle velocity becomes:
\[ v_p = 2\pi f Y_0 \cos\left(2\pi\left(ft - \frac{x}{\lambda}\right)\right) \]

### Step 4: Find maximum particle velocity (\(v_{p_{max}}\))
The maximum particle velocity occurs when the cosine function equals 1:
\[ v_{p_{max}} = 2\pi f Y_0 \]

### Step 5: 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:
\[ v_{p_{max}} = 4v \]
Substituting the expressions we derived:
\[ 2\pi f Y_0 = 4(\lambda f) \]

### Step 6: Simplify the equation
Cancel \(f\) from both sides (assuming \(f \neq 0\)):
\[ 2\pi Y_0 = 4\lambda \]
Rearranging gives:
\[ \lambda = \frac{2\pi Y_0}{4} = \frac{\pi Y_0}{2} \]

### Step 7: Identify the correct wavelength from the options
The derived wavelength \(\lambda = \frac{\pi Y_0}{2}\) corresponds to the second option given in the problem.

### Final Answer
The correct option is:
\[ \lambda = \frac{\pi Y_0}{2} \]

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