A transverse wave is represented by `y = A sin(omegat - kx)`.The wave velocity and the maximum particle velocity will be equal at the wavelength of

Options

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

Explanation

To solve the problem, we need to find the wavelength (λ) at which the wave velocity (Vw) and the maximum particle velocity (Vp_max) are equal for the given transverse wave represented by the equation:

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

### Step-by-Step Solution:

1. **Identify the Wave Velocity (Vw):**
The wave velocity (Vw) can be calculated using the relationship:
\[ V_w = \frac{\omega}{k} \]
where \( \omega \) is the angular frequency and \( k \) is the wave number.

2. **Determine the Maximum Particle Velocity (Vp_max):**
The particle velocity (Vp) is given by the time derivative of the displacement \( y \):
\[ V_p = \frac{dy}{dt} = A \omega \cos(\omega t - kx) \]
The maximum particle velocity (Vp_max) occurs when \( \cos(\omega t - kx) = 1 \):
\[ V_{p_{max}} = A \omega \]

3. **Set the Wave Velocity Equal to the Maximum Particle Velocity:**
We need to find the condition when these two velocities are equal:
\[ V_w = V_{p_{max}} \]
Substituting the expressions we found:
\[ \frac{\omega}{k} = A \omega \]

4. **Simplify the Equation:**
We can cancel \( \omega \) from both sides (assuming \( \omega \neq 0 \)):
\[ \frac{1}{k} = A \]

5. **Relate Wave Number (k) to Wavelength (λ):**
The wave number \( k \) is related to the wavelength \( \lambda \) by:
\[ k = \frac{2\pi}{\lambda} \]
Substituting this into the equation gives:
\[ \frac{1}{\frac{2\pi}{\lambda}} = A \]
Simplifying this:
\[ \frac{\lambda}{2\pi} = A \]

6. **Solve for Wavelength (λ):**
Rearranging the equation to find \( \lambda \):
\[ \lambda = 2\pi A \]

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