A particle is performing `S.H.M` with accerlration `a = 8 pi^(2) - 4 pi^(2) x` where `x` is coordinate of the particle `w.r.t` the origin. The parameters are in `S.I.` units. The particle is at rest at `x = -2` at `t = 0`.

Options

A. coordinate of the particle `w.r.t` origin at any time `t` is `2-4 cos 2pit`
B. coordinate of the particle `w.r.t` origin at any time `t` is `-2+4 sin 2pit`
C. coordinate of the particle `w.r.t` origin at any time `t` is `-4+2 cos 2pit`
D. the coordinate cannot be found because mas of the particle is not given.

Explanation

To solve the problem step by step, we will analyze the given acceleration equation and use the information provided about the particle's motion.

### Step 1: Write down the acceleration equation
The acceleration \( a \) of the particle is given by:
\[ a = 8\pi^2 - 4\pi^2 x \]

### Step 2: Relate acceleration to velocity
We know that acceleration can also be expressed in terms of velocity \( v \) and position \( x \):
\[ a = v \frac{dv}{dx} \]
Thus, we can equate the two expressions for acceleration:
\[ v \frac{dv}{dx} = 8\pi^2 - 4\pi^2 x \]

### Step 3: Separate variables
We can rearrange the equation to separate the variables:
\[ v \, dv = (8\pi^2 - 4\pi^2 x) \, dx \]

### Step 4: Integrate both sides
Now we will integrate both sides:
\[ \int v \, dv = \int (8\pi^2 - 4\pi^2 x) \, dx \]
This gives us:
\[ \frac{v^2}{2} = 8\pi^2 x - 2\pi^2 x^2 + C \]
where \( C \) is the constant of integration.

### Step 5: Solve for the constant of integration
We know that the particle is at rest at \( x = -2 \) when \( t = 0 \). This means that \( v = 0 \) at \( x = -2 \). Substituting these values into the equation:
\[ 0 = 8\pi^2(-2) - 2\pi^2(-2)^2 + C \]
Calculating this gives:
\[ 0 = -16\pi^2 - 8\pi^2 + C \implies C = 24\pi^2 \]

### Step 6: Substitute back to find the velocity equation
Now substituting \( C \) back into the equation:
\[ \frac{v^2}{2} = 8\pi^2 x - 2\pi^2 x^2 + 24\pi^2 \]
Multiplying through by 2 gives:
\[ v^2 = 16\pi^2 x - 4\pi^2 x^2 + 48\pi^2 \]

### Step 7: Express velocity in terms of \( x \)
We can factor out \( 4\pi^2 \):
\[ v^2 = 4\pi^2 (4x - x^2 + 12) \]
Taking the square root gives:
\[ v = 2\pi \sqrt{4x - x^2 + 12} \]

### Step 8: Relate velocity to time
Since \( v = \frac{dx}{dt} \), we can write:
\[ \frac{dx}{dt} = 2\pi \sqrt{4x - x^2 + 12} \]

### Step 9: Separate variables again
Rearranging gives:
\[ \frac{dx}{\sqrt{4x - x^2 + 12}} = 2\pi dt \]

### Step 10: Integrate both sides again
Now we will integrate both sides:
\[ \int \frac{dx}{\sqrt{4x - x^2 + 12}} = \int 2\pi dt \]
This integral can be simplified by completing the square in the expression under the square root.

### Step 11: Solve the integral
Completing the square for \( 4x - x^2 + 12 \):
\[ 4x - x^2 + 12 = - (x^2 - 4x - 12) = -((x - 2)^2 - 16) = 16 - (x - 2)^2 \]
Thus:
\[ \int \frac{dx}{\sqrt{16 - (x - 2)^2}} = 2\pi t + C \]
This integral results in:
\[ \sin^{-1}\left(\frac{x - 2}{4}\right) = 2\pi t + C \]

### Step 12: Find the constant \( C \)
At \( t = 0 \), \( x = -2 \):
\[ \sin^{-1}\left(\frac{-2 - 2}{4}\right) = C \implies \sin^{-1}(-1) = -\frac{\pi}{2} \implies C = -\frac{\pi}{2} \]

### Step 13: Final equation for \( x \)
Substituting \( C \) back gives:
\[ \sin^{-1}\left(\frac{x - 2}{4}\right) = 2\pi t - \frac{\pi}{2} \]
Taking the sine of both sides:
\[ \frac{x - 2}{4} = \sin\left(2\pi t - \frac{\pi}{2}\right) = -\cos(2\pi t) \]
Thus:
\[ x - 2 = -4\cos(2\pi t) \implies x = -4\cos(2\pi t) + 2 \]

### Final Answer
The position of the particle as a function of time is:
\[ x(t) = 2 - 4\cos(2\pi t) \]