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 function given by:

\[ y = A \sin^2(kx - \omega t + \phi) \]

### Step 1: Rewrite the wave equation using trigonometric identities

We can use the trigonometric identity for \(\sin^2\) to rewrite the wave equation. The identity states:

\[ \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \]

Applying this identity to our wave function:

\[ y = A \sin^2(kx - \omega t + \phi) = A \cdot \frac{1 - \cos(2(kx - \omega t + \phi))}{2} \]

This simplifies to:

\[ y = \frac{A}{2} (1 - \cos(2kx - 2\omega t + 2\phi)) \]

### Step 2: Identify the amplitude of the wave

From the rewritten equation, we can see that the amplitude of the wave is the coefficient of the cosine term. The term \( \frac{A}{2} \) indicates that the amplitude of the wave is:

\[ \text{Amplitude} = \frac{A}{2} \]

### Step 3: Determine the wavelength of the wave

The general form of a wave is given by:

\[ y = A \cos(kx - \omega t) \]

From this, we know that the wavelength \( \lambda \) is related to the wave number \( k \) by:

\[ \lambda = \frac{2\pi}{k} \]

In our case, the wave number \( k \) remains unchanged in the transformation we performed, so we can conclude that the wavelength is:

\[ \text{Wavelength} = \frac{2\pi}{k} \]

### Final Answer

Thus, the amplitude and wavelength of the wave are:

- Amplitude = \(\frac{A}{2}\)
- Wavelength = \(\frac{2\pi}{k}\)

The correct option is **C: Amplitude = \(\frac{A}{2}\) and Wavelength = \(\frac{2\pi}{k}\)**.

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