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 kinetic energy . `KE = E_(k) ( x, t)` at the point `(x, t)` is given by

Options

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

Explanation

To find the space density of the kinetic energy \( E_k(x, t) \) for the given longitudinal standing wave \( y = A \cos(kx) \cos(\omega t) \), we can follow these steps:

### Step 1: Understand the Kinetic Energy Expression
The kinetic energy density \( E_k \) for a wave can be expressed as:
\[ E_k = \frac{1}{2} \rho \left( \frac{dy}{dt} \right)^2 \]
where \( \rho \) is the density of the medium.

### Step 2: Differentiate the Wave Equation
We need to find \( \frac{dy}{dt} \) from the given wave equation. The wave equation is:
\[ y = A \cos(kx) \cos(\omega t) \]
To find \( \frac{dy}{dt} \), we differentiate \( y \) with respect to \( t \):
\[ \frac{dy}{dt} = A \cos(kx) \frac{d}{dt}(\cos(\omega t)) \]
Using the derivative of cosine, we have:
\[ \frac{d}{dt}(\cos(\omega t)) = -\omega \sin(\omega t) \]
Thus,
\[ \frac{dy}{dt} = A \cos(kx)(-\omega \sin(\omega t)) = -A \omega \cos(kx) \sin(\omega t) \]

### Step 3: Substitute into the Kinetic Energy Expression
Now we substitute \( \frac{dy}{dt} \) into the kinetic energy density formula:
\[ E_k = \frac{1}{2} \rho \left( -A \omega \cos(kx) \sin(\omega t) \right)^2 \]
This simplifies to:
\[ E_k = \frac{1}{2} \rho (A^2 \omega^2 \cos^2(kx) \sin^2(\omega t)) \]

### Step 4: Final Expression
Thus, we can express the kinetic energy density as:
\[ E_k(x, t) = \frac{1}{2} \rho A^2 \omega^2 \cos^2(kx) \sin^2(\omega t) \]

### Summary
The space density of the kinetic energy at the point \( (x, t) \) is given by:
\[ E_k(x, t) = \frac{1}{2} \rho A^2 \omega^2 \cos^2(kx) \sin^2(\omega t) \]