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 (Vmax) of a transverse wave represented by the equation \( y = A \sin(\omega t - kx) \).

### Step-by-step Solution:

1. **Identify the Maximum Particle Velocity (Vmax)**:
The maximum particle velocity for a wave can be derived from the wave equation. The maximum particle velocity is given by:
\[ V_{\text{max}} = A \cdot \omega \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

2. **Relate Angular Frequency to Wave Velocity and Wavelength**:
The angular frequency \( \omega \) can be expressed in terms of the wave velocity \( V \) and the wavelength \( \lambda \):
\[ \omega = 2\pi f \]
and since frequency \( f \) can be related to wave velocity and wavelength by:
\[ f = \frac{V}{\lambda} \]
we can substitute this into the equation for \( \omega \):
\[ \omega = 2\pi \left(\frac{V}{\lambda}\right) \]

3. **Substitute \( \omega \) into the Expression for Vmax**:
Now substituting \( \omega \) back into the expression for \( V_{\text{max}} \):
\[ V_{\text{max}} = A \cdot \left(2\pi \frac{V}{\lambda}\right) \]
This simplifies to:
\[ V_{\text{max}} = \frac{2\pi A V}{\lambda} \]

4. **Set Wave Velocity Equal to Maximum Particle Velocity**:
We are given that the wave velocity \( V \) is equal to the maximum particle velocity \( V_{\text{max}} \):
\[ V = \frac{2\pi A V}{\lambda} \]

5. **Solve for Wavelength (λ)**:
To find \( \lambda \), we can rearrange the equation:
\[ \lambda = 2\pi A \]

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