A transverse wave is described by the equation `y=A sin 2pi( nt- x//lambda_0)` . The maximum particle velocity is equal to 3 times the wave velocity if

Options

A. `lambda_0 = (piA)/(3) `
B. `lambda=(2pi A )/(3) `
C. `lambda_0 = pi A `
D. `lambda_0 = 3pi A `

Answer

To solve the problem, we need to analyze the given wave equation and relate the maximum particle velocity to the wave velocity. Let's break it down step by step.

### Step 1: Identify the parameters from the wave equation
The wave is described by the equation:
\[ y = A \sin(2\pi(nt - \frac{x}{\lambda_0})) \]
From this equation, we can identify:
- Amplitude \( A \)
- Angular frequency \( \omega = 2\pi n \)
- Wave number \( k = \frac{2\pi}{\lambda_0} \)

### Step 2: Write the expressions for maximum particle velocity and wave velocity
The maximum particle velocity \( v_p \) of a wave is given by:
\[ v_p = A \omega \]
The wave velocity \( v \) is given by:
\[ v = \frac{\omega}{k} \]

### Step 3: Relate maximum particle velocity to wave velocity
According to the problem, the maximum particle velocity is equal to 3 times the wave velocity:
\[ v_p = 3v \]
Substituting the expressions we found:
\[ A \omega = 3 \left( \frac{\omega}{k} \right) \]

### Step 4: Simplify the equation
We can cancel \( \omega \) from both sides (assuming \( \omega \neq 0 \)):
\[ A = \frac{3}{k} \]

### Step 5: Substitute the expression for \( k \)
We know that \( k = \frac{2\pi}{\lambda_0} \). Substituting this into the equation gives:
\[ A = \frac{3}{\frac{2\pi}{\lambda_0}} \]
This simplifies to:
\[ A = \frac{3 \lambda_0}{2\pi} \]

### Step 6: Rearranging to find \( \lambda_0 \)
Now, we can rearrange this equation to express \( \lambda_0 \):
\[ \lambda_0 = \frac{2\pi A}{3} \]

### Final Answer
Thus, the wavelength \( \lambda_0 \) is given by:
\[ \lambda_0 = \frac{2\pi A}{3} \]