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, we need to analyze the given wave equation \( y = A \sin^2(kx - \omega t) \) and determine its amplitude and frequency.

### Step-by-Step Solution:

1. **Identify the Given Equation**:
The wave equation provided is:
\[ y = A \sin^2(kx - \omega t) \]

2. **Use the Trigonometric Identity**:
We need to express \(\sin^2\) in terms of a standard sine function. We can use the trigonometric identity:
\[ \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \]
where \(\theta = kx - \omega t\).

3. **Rewrite the Equation**:
Applying the identity, we can rewrite the equation as:
\[ y = A \left(\frac{1 - \cos(2(kx - \omega t))}{2}\right) \]
Simplifying this gives:
\[ y = \frac{A}{2} (1 - \cos(2kx - 2\omega t)) \]

4. **Identify the Amplitude**:
The standard form of a wave equation is \(y = a \sin(kx - \omega t)\) or \(y = a \cos(kx - \omega t)\). In our modified equation, we can see that the amplitude of the wave is:
\[ \text{Amplitude} = \frac{A}{2} \]

5. **Identify the Frequency**:
The frequency \(f\) of a wave is given by the formula:
\[ f = \frac{\omega}{2\pi} \]
In our case, we have a cosine term with a coefficient of time as \(2\omega\). Therefore, the angular frequency \(\omega'\) is:
\[ \omega' = 2\omega \]
Thus, the frequency becomes:
\[ f = \frac{2\omega}{2\pi} = \frac{\omega}{\pi} \]

6. **Final Answers**:
- Amplitude: \(\frac{A}{2}\)
- Frequency: \(\frac{\omega}{\pi}\)

### Summary of Results:
- Amplitude = \(\frac{A}{2}\)
- Frequency = \(\frac{\omega}{\pi}\)