The equation y=Asin^2(kx-omegat)` represents a wave motion with

Options

A. amplitude A, frequency `omega/2pi`
B. amplitude A/2, frequency `omega/pi`
C. amplitude 2A, frequencey `omega/4pi`
D. does not represent a wave motion

Answer

To solve the problem regarding the wave motion represented by the equation \( y = A \sin^2(kx - \omega t) \), we will follow these steps:

### Step 1: Rewrite the Sine Squared Function
The first step is to rewrite the sine squared term using a trigonometric identity. The identity states that:
\[ \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \]
Applying this identity to our equation, we have:
\[ y = A \sin^2(kx - \omega t) = A \cdot \frac{1 - \cos(2(kx - \omega t))}{2} \]

### Step 2: Simplify the Equation
Now, we can simplify the equation:
\[ y = \frac{A}{2} (1 - \cos(2(kx - \omega t))) \]
This can be rewritten as:
\[ y = \frac{A}{2} - \frac{A}{2} \cos(2kx - 2\omega t) \]

### Step 3: Identify the Amplitude
In the standard wave equation form \( y = C - D \cos(Bx - Ct) \), the amplitude is given by the coefficient of the cosine term. Here, the amplitude \( A' \) is:
\[ A' = \frac{A}{2} \]

### Step 4: Identify the Angular Frequency
The angular frequency \( \omega' \) in the cosine term \( \cos(2kx - 2\omega t) \) is \( 2\omega \).

### Step 5: Relate Angular Frequency to Frequency
The relationship between angular frequency \( \omega \) and frequency \( f \) is given by:
\[ \omega = 2\pi f \]
Thus, for \( 2\omega \):
\[ 2\omega = 2\pi f' \]
From this, we can derive the frequency \( f' \):
\[ f' = \frac{2\omega}{2\pi} = \frac{\omega}{\pi} \]

### Final Results
- The amplitude of the wave is \( \frac{A}{2} \).
- The frequency of the wave is \( \frac{\omega}{\pi} \).