Simple harmonic wave is represented 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. `2pi a_(0)//3`
C. `pi a_(0)`
D. `pi a_(0)//2`

Explanation

To solve the problem, we start with the given wave equation:

\[ y(x, t) = a_0 \sin\left(2\pi\left(vt - \frac{x}{\lambda}\right)\right) \]

### Step 1: Find the expression for maximum particle velocity

The maximum particle velocity \( v_p \) can be found by differentiating the displacement \( y \) with respect to time \( t \):

\[ v_p = \frac{dy}{dt} = a_0 \cdot \frac{d}{dt}\left(\sin\left(2\pi\left(vt - \frac{x}{\lambda}\right)\right)\right) \]

Using the chain rule, we get:

\[ v_p = a_0 \cdot \cos\left(2\pi\left(vt - \frac{x}{\lambda}\right)\right) \cdot \frac{d}{dt}\left(2\pi\left(vt - \frac{x}{\lambda}\right)\right) \]

The derivative of \( 2\pi(vt - \frac{x}{\lambda}) \) with respect to \( t \) is \( 2\pi v \). Therefore:

\[ v_p = a_0 \cdot \cos\left(2\pi\left(vt - \frac{x}{\lambda}\right)\right) \cdot 2\pi v \]

The maximum value of \( \cos \) is 1, so:

\[ v_{p, \text{max}} = a_0 \cdot 2\pi v \]

### Step 2: Relate maximum particle velocity to wave velocity

We are given that the maximum particle velocity is three times the wave velocity \( v \):

\[ v_{p, \text{max}} = 3v \]

Substituting the expression we found for \( v_{p, \text{max}} \):

\[ a_0 \cdot 2\pi v = 3v \]

### Step 3: Solve for \( a_0 \)

To isolate \( a_0 \), we can divide both sides by \( v \) (assuming \( v \neq 0 \)):

\[ a_0 \cdot 2\pi = 3 \]

Now, solving for \( a_0 \):

\[ a_0 = \frac{3}{2\pi} \]

### Step 4: Find the wave velocity in terms of wavelength

The wave velocity \( v \) can also be expressed in terms of angular frequency \( \omega \) and wave number \( k \):

\[ v = \frac{\omega}{k} \]

Where:

\[ \omega = 2\pi v \quad \text{and} \quad k = \frac{2\pi}{\lambda} \]

Thus, we can substitute:

\[ v = \frac{2\pi v}{\frac{2\pi}{\lambda}} = v \cdot \lambda \]

### Step 5: Set up the equation

From our earlier equation \( a_0 \cdot 2\pi v = 3v \), we can substitute \( a_0 \):

\[ \frac{3}{2\pi} \cdot 2\pi v = 3v \]

### Step 6: Solve for \( \lambda \)

Since we know \( v = v \), we can equate the expressions:

\[ \lambda = \frac{2\pi}{3} \cdot a_0 \]

Substituting \( a_0 = \frac{3}{2\pi} \):

\[ \lambda = \frac{2\pi}{3} \cdot \frac{3}{2\pi} = 1 \]

Thus, the wavelength \( \lambda \) of the wave is:

\[ \lambda = 1 \]

### Final Answer:

The wavelength \( \lambda \) of the wave is \( 1 \).

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