A transverse wave is represented by `y=Asin(omegat-kx)`. For what value of the wavelength is the wave velocity equal to the maximum particle velocity?

Options

A. `(piA)/2`
B. `piA`
C. `2piA`
D. A

Explanation

To solve the problem, we need to find the value of the wavelength (λ) for which the wave velocity (v) is equal to the maximum particle velocity (v_max) of a transverse wave represented by the equation:

\[ y = A \sin(\omega t - kx) \]

### Step-by-Step Solution:

1. **Identify the wave equation parameters**:
- The wave equation is given as \( y = A \sin(\omega t - kx) \).
- Here, \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( k \) is the wave number.

2. **Determine the wave velocity (v)**:
- The wave velocity \( v \) is given by the formula:
\[ v = \frac{\omega}{k} \]

3. **Determine the particle velocity (v_p)**:
- The particle velocity is the derivative of displacement with respect to time:
\[ v_p = \frac{\partial y}{\partial t} = A \omega \cos(\omega t - kx) \]
- The maximum particle velocity \( v_{max} \) occurs when \( \cos(\omega t - kx) = 1 \):
\[ v_{max} = A \omega \]

4. **Set the wave velocity equal to the maximum particle velocity**:
- According to the problem, we need to find when:
\[ v_{max} = v \]
- Substituting the expressions we have:
\[ A \omega = \frac{\omega}{k} \]

5. **Cancel out \( \omega \) (assuming \( \omega \neq 0 \))**:
- Dividing both sides by \( \omega \):
\[ A = \frac{1}{k} \]

6. **Substitute the wave number \( k \)**:
- Recall that the wave number \( k \) is related to the wavelength \( \lambda \) by:
\[ k = \frac{2\pi}{\lambda} \]
- Substitute this into the equation:
\[ A = \frac{1}{\frac{2\pi}{\lambda}} = \frac{\lambda}{2\pi} \]

7. **Rearranging to find the wavelength**:
- Rearranging gives:
\[ \lambda = 2\pi A \]

### Final Answer:
The wavelength \( \lambda \) for which the wave velocity is equal to the maximum particle velocity is:
\[ \lambda = 2\pi A \]