A simple harmonic wave is represent by the relation
`y(x,t)=a_(0) sin 2pi(vt-(x)/(lambda))`
if the maximum particle velocity is three times the wave velocity, the wavelength `lambda` of the wave is

Options

A. `pi a_(0)//3`
B. `2pia_(0)//3`
C. `pia_(0)`
D. `pi a_(0)//2`

Explanation

To solve the problem, we need to find the wavelength \( \lambda \) of a simple harmonic wave given that the maximum particle velocity is three times the wave velocity. Let's break down the solution step by step.

### Step 1: Understand the wave equation
The wave is represented by the equation:
\[ y(x,t) = a_0 \sin(2\pi(vt - \frac{x}{\lambda})) \]
where:
- \( a_0 \) is the amplitude,
- \( v \) is the wave velocity,
- \( \lambda \) is the wavelength.

### Step 2: Identify the maximum particle velocity
The maximum particle velocity \( v_p \) for a wave is given by the formula:
\[ v_p = \omega a_0 \]
where \( \omega \) is the angular frequency.

### Step 3: Relate angular frequency to frequency
The angular frequency \( \omega \) is related to the frequency \( f \) by:
\[ \omega = 2\pi f \]
Thus, we can express the maximum particle velocity as:
\[ v_p = 2\pi f a_0 \]

### Step 4: Relate wave velocity to frequency and wavelength
The wave velocity \( v \) is related to frequency and wavelength by:
\[ v = f \lambda \]

### Step 5: Set up the equation based on the problem statement
According to the problem, the maximum particle velocity is three times the wave velocity:
\[ v_p = 3v \]
Substituting the expressions we have:
\[ 2\pi f a_0 = 3(f \lambda) \]

### Step 6: Simplify the equation
We can cancel \( f \) from both sides (assuming \( f \neq 0 \)):
\[ 2\pi a_0 = 3\lambda \]

### Step 7: Solve for wavelength \( \lambda \)
Rearranging the equation gives:
\[ \lambda = \frac{2\pi a_0}{3} \]

### Final Answer
Thus, the wavelength \( \lambda \) of the wave is:
\[ \lambda = \frac{2\pi a_0}{3} \]

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