A longitudinal standing wave ` y = a cos kx cos omega t` is maintained in a homogeneious medium of density `rho`. Here `omega` is the angular speed and `k` , the wave number and `a` is the amplitude of the standing wave . This standing wave exists all over a given region of space.
The space density of the potential energy `PE = E_(p)(x , t)` at a point `(x , t)` in this space is

Options

A. `E_(p) = ( rho a^(2) omega^(2))/(2)`
B. `E_(p) = ( rho a^(2) omega^(2))/(2) cos^(2) kx sin^(2) omega t`
C. `E_(p) = ( rho a^(2) omega^(2))/(2) sin^(2) kx cos^(2) omega t`
D. `E_(p) = ( rho a^(2) omega^(2))/(2) sin^(2) kx sin^(2) omega t`

Explanation

To find the space density of the potential energy \( PE = E_p(x, t) \) at a point \( (x, t) \) in the medium for the given longitudinal standing wave \( y = a \cos(kx) \cos(\omega t) \), we will follow these steps:

### Step 1: Understand the Potential Energy Formula
The potential energy density \( E_p \) in a medium can be expressed as:
\[ E_p = \frac{1}{2} E \left( \frac{dy}{dx} \right)^2 \]
where \( E \) is the Young's modulus of the medium and \( \frac{dy}{dx} \) is the spatial derivative of the wave function.

### Step 2: Differentiate the Wave Function
We need to find \( \frac{dy}{dx} \) from the wave function:
\[ y = a \cos(kx) \cos(\omega t) \]
Differentiating \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = -a k \sin(kx) \cos(\omega t) \]

### Step 3: Substitute into the Potential Energy Formula
Now, we substitute \( \frac{dy}{dx} \) into the potential energy density formula:
\[ E_p = \frac{1}{2} E \left( -a k \sin(kx) \cos(\omega t) \right)^2 \]
This simplifies to:
\[ E_p = \frac{1}{2} E (a^2 k^2 \sin^2(kx) \cos^2(\omega t)) \]

### Step 4: Relate Young's Modulus to Density and Wave Velocity
From the wave mechanics, we know that the wave velocity \( V \) in the medium is given by:
\[ V = \sqrt{\frac{E}{\rho}} \]
Thus, we can express \( E \) as:
\[ E = \rho V^2 \]

### Step 5: Substitute \( E \) into the Potential Energy Expression
Substituting \( E \) back into the expression for \( E_p \):
\[ E_p = \frac{1}{2} (\rho V^2) (a^2 k^2 \sin^2(kx) \cos^2(\omega t)) \]

### Step 6: Substitute \( k \) in Terms of \( \omega \) and \( V \)
We know that the wave number \( k \) is related to the angular frequency \( \omega \) and wave velocity \( V \) by:
\[ k = \frac{\omega}{V} \]
Substituting \( k \) into the expression for \( E_p \):
\[ E_p = \frac{1}{2} \rho V^2 \left( a^2 \left( \frac{\omega}{V} \right)^2 \sin^2(kx) \cos^2(\omega t) \right) \]
This simplifies to:
\[ E_p = \frac{1}{2} \rho \frac{\omega^2 a^2}{V^2} V^2 \sin^2(kx) \cos^2(\omega t) \]
Thus, we have:
\[ E_p = \frac{1}{2} \rho \omega^2 a^2 \sin^2(kx) \cos^2(\omega t) \]

### Final Expression
The final expression for the space density of the potential energy \( E_p(x, t) \) is:
\[ E_p(x, t) = \frac{1}{2} \rho \omega^2 a^2 \sin^2(kx) \cos^2(\omega t) \]

### Conclusion
Thus, the answer is:
\[ E_p(x, t) = \frac{1}{2} \rho a^2 \omega^2 \sin^2(kx) \cos^2(\omega t) \]