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.
If a graph `E ( = E_(p) + E_(k))` versus `t` , i.e., total space energy density verus time were drawn at the instants of time `t = 0` and `t = T//4`, between two successive nodes separated by distance `lambda//2` which of the following graphs correctly shows the total energy `(E)` distribution at the two instants.

Options

A.
B.
C.
D.

Explanation

To solve the problem of finding the correct graph for the total energy distribution of a longitudinal standing wave at two different time instances, we will follow these steps:

### Step 1: Understand the standing wave equation
The given standing wave is represented by the equation:
\[ y = a \cos(kx) \cos(\omega t) \]
where:
- \( a \) is the amplitude,
- \( k \) is the wave number,
- \( \omega \) is the angular frequency.

### Step 2: Write the expression for total energy density
The total energy density \( E \) is the sum of potential energy density \( E_p \) and kinetic energy density \( E_k \):
\[ E = E_p + E_k \]
The expressions for potential and kinetic energy densities are:
\[ E_p = \frac{1}{2} Y \left( \frac{dy}{dx} \right)^2 \]
\[ E_k = \frac{1}{2} \rho \left( \frac{dy}{dt} \right)^2 \]
where \( Y \) is Young's modulus and \( \rho \) is the density of the medium.

### Step 3: Calculate \( E \) at \( t = 0 \)
At \( t = 0 \):
\[ y = a \cos(kx) \cos(0) = a \cos(kx) \]
Now, we differentiate \( y \) with respect to \( x \) and \( t \):
1. \( \frac{dy}{dx} = -a k \sin(kx) \)
2. \( \frac{dy}{dt} = -a \omega \sin(\omega t) \)

Substituting these into the energy density equations:
\[ E_p = \frac{1}{2} Y \left(-a k \sin(kx)\right)^2 = \frac{1}{2} Y a^2 k^2 \sin^2(kx) \]
\[ E_k = \frac{1}{2} \rho \left(-a \omega \sin(0)\right)^2 = 0 \]

Thus, at \( t = 0 \):
\[ E = E_p = \frac{1}{2} Y a^2 k^2 \sin^2(kx) \]

### Step 4: Calculate \( E \) at \( t = \frac{T}{4} \)
At \( t = \frac{T}{4} \):
\[ y = a \cos(kx) \cos\left(\frac{\pi}{2}\right) = 0 \]
Now, substituting \( t = \frac{T}{4} \) into the derivatives:
1. \( \frac{dy}{dx} = -a k \sin(kx) \)
2. \( \frac{dy}{dt} = -a \omega \sin\left(\frac{\pi}{2}\right) = -a \omega \)

Now substituting into the energy density equations:
\[ E_p = 0 \]
\[ E_k = \frac{1}{2} \rho \left(-a \omega\right)^2 = \frac{1}{2} \rho a^2 \omega^2 \]

Thus, at \( t = \frac{T}{4} \):
\[ E = E_k = \frac{1}{2} \rho a^2 \omega^2 \]

### Step 5: Analyze the graphs
1. At \( t = 0 \), the energy density \( E \) is proportional to \( \sin^2(kx) \), which has nodes at \( x = n\frac{\pi}{k} \) (where \( n \) is an integer).
2. At \( t = \frac{T}{4} \), the energy density \( E \) is proportional to \( \cos^2(kx) \), which has nodes at \( x = (n + \frac{1}{2})\frac{\pi}{k} \).

### Step 6: Identify the correct graph
From the analysis, the graph for \( E \) at \( t = 0 \) should show peaks at \( x = \frac{\pi}{2k} \) and \( x = \frac{3\pi}{2k} \) (sine squared function), while at \( t = \frac{T}{4} \), the graph should show peaks at \( x = 0 \) and \( x = \pi \) (cosine squared function).

Based on the options provided, the correct graph that represents these conditions is **Option A**.