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 find the condition under which the maximum particle velocity of a transverse wave is equal to three times the wave velocity. The wave is described by the equation:

\[ y = A \sin\left(2\pi\left(nt - \frac{x}{\lambda_0}\right)\right) \]

### Step-by-Step Solution:

1. **Identify the parameters from the wave equation**:
- Amplitude \( A \)
- Angular frequency \( \omega = 2\pi n \)
- Wave number \( k = \frac{2\pi}{\lambda_0} \)

2. **Write the expression for maximum particle velocity**:
The maximum particle velocity \( v_{max} \) is given by:
\[ v_{max} = A \omega \]

3. **Write the expression for wave velocity**:
The wave velocity \( V \) is given by:
\[ V = \frac{\omega}{k} \]

4. **Substitute the expressions for \( \omega \) and \( k \)**:
From the wave number, we have:
\[ k = \frac{2\pi}{\lambda_0} \]
Thus, the wave velocity becomes:
\[ V = \frac{\omega}{k} = \frac{2\pi n}{\frac{2\pi}{\lambda_0}} = n \lambda_0 \]

5. **Set up the equation based on the given condition**:
According to the problem, the maximum particle velocity is three times the wave velocity:
\[ A \omega = 3V \]
Substituting for \( V \):
\[ A \omega = 3(n \lambda_0) \]

6. **Substitute \( \omega = 2\pi n \) into the equation**:
\[ A (2\pi n) = 3(n \lambda_0) \]

7. **Cancel \( n \) from both sides (assuming \( n \neq 0 \))**:
\[ 2\pi A = 3\lambda_0 \]

8. **Solve for \( \lambda_0 \)**:
\[ \lambda_0 = \frac{2\pi A}{3} \]

### Final Result:
The wavelength \( \lambda_0 \) is given by:
\[ \lambda_0 = \frac{2\pi A}{3} \]