A wave is represente by `y = Asin^2 (kx - omegat + phi)`. The amplitude and wavelength of wave is given by

Options

A. `2 A, (2pi)/k`
B. `A, (2pi)/k`
C. `A/2, (2pi)/k`
D. `A/2, (pi)/k`

Answer

To solve the problem, we need to analyze the wave equation given as \( y = A \sin^2(kx - \omega t + \phi) \) and determine its amplitude and wavelength.

### Step 1: Rewrite the Wave Equation
We start with the equation:
\[ y = A \sin^2(kx - \omega t + \phi) \]
Using the trigonometric identity for \(\sin^2\), we can rewrite \(\sin^2 \theta\) as:
\[ \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \]
Here, \(\theta = kx - \omega t + \phi\). Therefore, we can express \(\sin^2(kx - \omega t + \phi)\) as:
\[ \sin^2(kx - \omega t + \phi) = \frac{1 - \cos(2(kx - \omega t + \phi))}{2} \]

### Step 2: Substitute Back into the Equation
Substituting this back into the wave equation gives:
\[ y = A \left(\frac{1 - \cos(2(kx - \omega t + \phi))}{2}\right) \]
Simplifying this, we get:
\[ y = \frac{A}{2} - \frac{A}{2} \cos(2(kx - \omega t + \phi)) \]

### Step 3: Identify Amplitude
In the rewritten equation, the term \(-\frac{A}{2} \cos(2(kx - \omega t + \phi))\) indicates that the amplitude of the wave is the coefficient of the cosine term. Therefore, the amplitude \(A_{\text{wave}}\) is:
\[ A_{\text{wave}} = \frac{A}{2} \]

### Step 4: Determine the Wavelength
The wave number \(k\) is related to the wavelength \(\lambda\) by the formula:
\[ k = \frac{2\pi}{\lambda} \]
In our case, the argument of the cosine function is \(2(kx - \omega t + \phi)\), which implies that the wave number is \(2k\). Therefore, we can express the wavelength as:
\[ 2k = \frac{2\pi}{\lambda} \implies \lambda = \frac{2\pi}{2k} = \frac{\pi}{k} \]

### Final Results
Thus, the amplitude and wavelength of the wave are:
- Amplitude: \( \frac{A}{2} \)
- Wavelength: \( \frac{\pi}{k} \)

### Summary of Answers
- Amplitude: \( \frac{A}{2} \)
- Wavelength: \( \frac{\pi}{k} \)