`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 find the amplitude \( A' \) and frequency \( f \) of the wave represented by the equation \( Y = 2A \cos^2(kx - \omega t) \), we can follow these steps:

### Step 1: Rewrite the given equation
The given wave equation is:
\[ Y = 2A \cos^2(kx - \omega t) \]
We know from trigonometric identities that:
\[ \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} \]
Using this identity, we can rewrite the equation.

### Step 2: Apply the trigonometric identity
Substituting \( \theta = kx - \omega t \) into the identity:
\[ \cos^2(kx - \omega t) = \frac{1 + \cos(2(kx - \omega t))}{2} \]
Thus, we can rewrite \( Y \):
\[ Y = 2A \left(\frac{1 + \cos(2(kx - \omega t))}{2}\right) \]
This simplifies to:
\[ Y = A(1 + \cos(2(kx - \omega t))) \]

### Step 3: Separate the terms
Now, we can express \( Y \) as:
\[ Y = A + A \cos(2(kx - \omega t)) \]
This indicates that the wave has a constant term \( A \) and an oscillatory term \( A \cos(2(kx - \omega t)) \).

### Step 4: Identify the amplitude \( A' \)
From the oscillatory term \( A \cos(2(kx - \omega t)) \), we can see that the amplitude \( A' \) of the wave is:
\[ A' = A \]

### Step 5: Identify the frequency \( f \)
The angular frequency \( \omega \) is related to the frequency \( f \) by the equation:
\[ \omega = 2\pi f \]
From our equation, we can see that the angular frequency for the oscillatory term is \( 2\omega \). Thus, we can write:
\[ 2\omega = 2\pi f' \]
This implies:
\[ f' = \frac{\omega}{\pi} \]

### Final Result
Thus, the values of the amplitude \( A' \) and frequency \( f \) are:
\[ A' = A \quad \text{and} \quad f = \frac{\omega}{2\pi} \]