A transverse wave is represented by the equation
`y=y_0sin.(2pi)/(lamda)(vt-x)`
For what value of `lamda`, the maximum particle velocity equal to two times the wave velocity?
(A) `lamda= 2piy_(0)` (B) `lamda= (piy_(0))/3` (C) `lamda= (piy_(0))/2` (D) `lamda= piy_(0)`

Options

A. `lamda= 2piy_(0)`
B. `lamda= (piy_(0))/3`
C. `lamda= (piy_(0))/2`
D. `lamda= piy_(0)`

Answer

To solve the problem, we need to determine the value of \( \lambda \) for which the maximum particle velocity equals two times the wave velocity. Let's break it down step by step.

### Step 1: Understand the wave equation
The given wave equation is:
\[ y = y_0 \sin\left(\frac{2\pi}{\lambda}(vt - x)\right) \]
Here, \( y_0 \) is the amplitude, \( \lambda \) is the wavelength, \( v \) is the wave velocity, and \( x \) is the position.

### Step 2: Find the maximum particle velocity
The maximum particle velocity \( v_p \) for a transverse wave is given by:
\[ v_p = A \omega \]
where \( A \) is the amplitude (which is \( y_0 \) in this case) and \( \omega \) is the angular frequency.

### Step 3: Relate angular frequency to wave velocity
The angular frequency \( \omega \) can be expressed in terms of frequency \( f \):
\[ \omega = 2\pi f \]
The wave velocity \( v \) is related to frequency and wavelength by:
\[ v = f \lambda \implies f = \frac{v}{\lambda} \]
Substituting \( f \) into the expression for \( \omega \):
\[ \omega = 2\pi \left(\frac{v}{\lambda}\right) \]

### Step 4: Substitute \( \omega \) into the maximum particle velocity equation
Now, substituting \( \omega \) into the maximum particle velocity equation:
\[ v_p = y_0 \cdot \omega = y_0 \cdot 2\pi \left(\frac{v}{\lambda}\right) \]
Thus, we have:
\[ v_p = \frac{2\pi y_0 v}{\lambda} \]

### Step 5: Set the condition for maximum particle velocity
According to the problem, we need to set the maximum particle velocity equal to two times the wave velocity:
\[ v_p = 2v \]
Substituting the expression we found for \( v_p \):
\[ \frac{2\pi y_0 v}{\lambda} = 2v \]

### Step 6: Solve for \( \lambda \)
We can cancel \( v \) from both sides (assuming \( v \neq 0 \)):
\[ \frac{2\pi y_0}{\lambda} = 2 \]
Now, rearranging gives:
\[ \lambda = \pi y_0 \]

### Conclusion
Thus, the value of \( \lambda \) for which the maximum particle velocity equals two times the wave velocity is:
\[ \lambda = \pi y_0 \]
The correct answer is option (D) \( \lambda = \pi y_0 \).