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 derive the position function of the particle performing Simple Harmonic Motion (SHM).

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

### Step 2: Rewrite the acceleration equation
We can factor out \( 4\pi^2 \) from the acceleration equation:
\[ a = 4\pi^2 (2 - x) \]

### Step 3: Relate acceleration to SHM
In SHM, the acceleration \( a \) can also be expressed as:
\[ a = -\omega^2 x \]
where \( \omega \) is the angular frequency.

### Step 4: Identify the mean position
From the equation \( a = 4\pi^2 (2 - x) \), we can see that the acceleration is zero when \( x = 2 \). This indicates that \( x = 2 \) is the mean position of the SHM.

### Step 5: Determine the extreme position
The particle is at rest at \( x = -2 \) at \( t = 0 \). Since it is at rest at this position, \( x = -2 \) is an extreme position of the motion.

### Step 6: Calculate the amplitude
The amplitude \( A \) of the SHM can be calculated as the distance from the mean position to the extreme position:
\[ A = |2 - (-2)| = 4 \]

### Step 7: Write the general equation of motion
The general equation for the position \( x(t) \) in SHM can be expressed as:
\[ x(t) = x_m + A \cos(\omega t + \phi) \]
where \( x_m \) is the mean position, \( A \) is the amplitude, and \( \phi \) is the phase constant.

### Step 8: Substitute known values
Here, \( x_m = 2 \) and \( A = 4 \). We need to determine \( \omega \) and \( \phi \). From the acceleration equation, we can find \( \omega^2 = 4\pi^2 \), thus:
\[ \omega = 2\pi \]

### Step 9: Determine the phase constant \( \phi \)
At \( t = 0 \), the particle is at \( x = -2 \):
\[ -2 = 2 + 4 \cos(0 + \phi) \]
This simplifies to:
\[ -2 = 2 + 4 \cos(\phi) \]
\[ -4 = 4 \cos(\phi) \]
\[ \cos(\phi) = -1 \]
This implies \( \phi = \pi \).

### Step 10: Write the final equation of motion
Substituting \( \omega \) and \( \phi \) into the equation of motion:
\[ x(t) = 2 + 4 \cos(2\pi t + \pi) \]
Using the property of cosine, \( \cos(\theta + \pi) = -\cos(\theta) \):
\[ x(t) = 2 - 4 \cos(2\pi t) \]

### Final Result
Thus, the coordinate of the particle at any time \( t \) is:
\[ x(t) = 2 - 4 \cos(2\pi t) \]

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