`Y= 2A cos^(2) (kx - omega t)` represents a wave with amplitude A´ and frequency f. The value of A´ and f are:

Options

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

Explanation

To solve the problem, we need to analyze the wave equation given by:

\[ Y = 2A \cos^2(kx - \omega t) \]

### Step 1: Rewrite the wave equation using a trigonometric identity
We can use the trigonometric identity for cosine squared:

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

Applying this identity to our wave equation:

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

### Step 2: Simplify the equation
This simplifies to:

\[ Y = A(1 + \cos(2(kx - \omega t))) \]

### Step 3: Further simplify the equation
Distributing \( A \):

\[ Y = A + A \cos(2(kx - \omega t)) \]

### Step 4: Identify the amplitude and frequency
In the form \( Y = A + B \cos(Cx - Dt) \), we can identify:

- The amplitude \( A' \) of the wave is the coefficient of the cosine term, which is \( A \).
- The angular frequency \( \omega' \) is \( 2\omega \) (from the term \( \cos(2(kx - \omega t)) \)).

### Step 5: Relate angular frequency to frequency
The relationship between angular frequency \( \omega' \) and frequency \( f' \) is given by:

\[ \omega' = 2\pi f' \]

Setting \( \omega' = 2\omega \):

\[ 2\omega = 2\pi f' \]

Dividing both sides by 2:

\[ \omega = \pi f' \]

### Step 6: Solve for frequency
Thus, we can express the frequency \( f' \) as:

\[ f' = \frac{\omega}{\pi} \]

### Conclusion
From our analysis, we find:
- The amplitude \( A' \) is \( A \).
- The frequency \( f' \) is \( \frac{\omega}{\pi} \).

### Final Answer
- Amplitude \( A' = A \)
- Frequency \( f' = \frac{\omega}{\pi} \)

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