A transverse wave is given by `y=A sin 2 pi ((t)/(T)-(x)/(lambda))`. The maximum particle velocity is equal to 4 times the wave velocity when

Options

A. `lambda = 2 pi A`
B. ` lambda=(1)/(2) pi A`
C. ` lambda = pi A`
D. `lambda =(1)/(4) pi A`

Answer

To solve the problem, we need to analyze the given transverse wave equation and derive the necessary expressions for maximum particle velocity and wave velocity.

### Step-by-Step Solution:

1. **Identify the wave equation**:
The given wave equation is:
\[ y = A \sin\left(2\pi\left(\frac{t}{T} - \frac{x}{\lambda}\right)\right) \]
where \( A \) is the amplitude, \( T \) is the time period, and \( \lambda \) is the wavelength.

2. **Find the maximum particle velocity**:
The particle velocity \( v_p \) is given by the time derivative of the displacement \( y \):
\[ v_p = \frac{dy}{dt} = A \cdot \frac{d}{dt}\left(\sin\left(2\pi\left(\frac{t}{T} - \frac{x}{\lambda}\right)\right)\right) \]
Using the chain rule:
\[ v_p = A \cdot \cos\left(2\pi\left(\frac{t}{T} - \frac{x}{\lambda}\right)\right) \cdot \frac{d}{dt}\left(2\pi\left(\frac{t}{T} - \frac{x}{\lambda}\right)\right) \]
\[ = A \cdot \cos\left(2\pi\left(\frac{t}{T} - \frac{x}{\lambda}\right)\right) \cdot \left(\frac{2\pi}{T}\right) \]
Therefore, the expression for particle velocity becomes:
\[ v_p = \frac{2\pi A}{T} \cos\left(2\pi\left(\frac{t}{T} - \frac{x}{\lambda}\right)\right) \]

3. **Determine the maximum particle velocity**:
The maximum value of \( \cos \) is 1, hence:
\[ v_{p,\text{max}} = \frac{2\pi A}{T} \]

4. **Find the wave velocity**:
The wave velocity \( v_w \) is given by:
\[ v_w = \frac{\lambda}{T} \]

5. **Set up the equation for the condition given in the problem**:
We know from the problem statement that the maximum particle velocity is equal to 4 times the wave velocity:
\[ v_{p,\text{max}} = 4 v_w \]
Substituting the expressions we derived:
\[ \frac{2\pi A}{T} = 4 \left(\frac{\lambda}{T}\right) \]

6. **Simplify the equation**:
Cancel \( T \) from both sides (assuming \( T \neq 0 \)):
\[ 2\pi A = 4\lambda \]
Rearranging gives:
\[ \lambda = \frac{\pi A}{2} \]

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