Consider a system of two equal points charges, each `Q = 8 muC`, which are fixed at points (2m, 0 ) and (-2m, 0). Another charge `mu` is held at a point (0,0.1m) on the y-axis. Mass of the charge `mu` is 91 mg . At t= 0 , `mu` is released from rest and it is observed to oscillate along y-axis in a simple harmonic manner. It is also observed that at t = 0 , the force experienced by it is ` 9 xx 10^(-3)N`.
Equation of SHM (displacement from mean position) can be expressed as

Options

A. `y = 0.1 sin (10pit)`
B. `y = 0.1sin(10pit + pi//2)`
C. `y = 0.1 sin (5pit+ pi//2)`
D. `y = 0.2 sin (5pit)`

Explanation

To solve the problem step by step, we will analyze the forces acting on the charge `μ` and derive the equation of motion for its simple harmonic motion (SHM).

### Step 1: Understand the System
We have two equal point charges, each `Q = 8 μC`, located at coordinates (2m, 0) and (-2m, 0). A charge `μ` with a mass of 91 mg is placed at (0, 0.1m) on the y-axis. The system is symmetric about the y-axis.

### Step 2: Calculate the Force on Charge `μ`
The force acting on charge `μ` due to the two point charges can be calculated using Coulomb's law. The force experienced by `μ` at the position (0, 0.1m) is given as `F = 9 × 10^(-3) N`.

### Step 3: Calculate the Acceleration of Charge `μ`
Using Newton's second law, we can find the acceleration `a` of charge `μ`:
\[ a = \frac{F}{m} \]
Where:
- \( F = 9 \times 10^{-3} \, N \)
- \( m = 91 \, mg = 91 \times 10^{-6} \, kg \)

Substituting the values:
\[ a = \frac{9 \times 10^{-3}}{91 \times 10^{-6}} \]
Calculating this gives:
\[ a ≈ 98.9 \, m/s^2 \]

### Step 4: Determine the Displacement from the Mean Position
The mean position is at (0, 0). The displacement from the mean position is given as:
\[ y = 0.1 \, m \]

### Step 5: Calculate the Angular Frequency (ω)
The angular frequency \( \omega \) for SHM can be calculated using the formula:
\[ \omega = \sqrt{\frac{a}{y}} \]
Substituting the values:
\[ \omega = \sqrt{\frac{98.9}{0.1}} \]
Calculating this gives:
\[ \omega ≈ 31.44 \, rad/s \]

### Step 6: Express Angular Frequency in Terms of π
To express \( \omega \) in terms of π:
\[ 31.44 \, rad/s \approx 10\pi \, rad/s \]

### Step 7: Write the Equation of SHM
The general equation for SHM is given by:
\[ y(t) = A \sin(\omega t + \phi) \]
Where:
- \( A = 0.1 \, m \) (amplitude)
- \( \omega = 10\pi \, rad/s \)
- \( \phi \) is the phase constant.

### Step 8: Determine the Phase Constant (φ)
At \( t = 0 \), the position of the charge is \( y(0) = 0.1 \, m \):
\[ 0.1 = 0.1 \sin(0 + \phi) \]
This simplifies to:
\[ \sin(\phi) = 1 \]
Thus, \( \phi = \frac{\pi}{2} \).

### Final Equation of SHM
Substituting the values into the SHM equation:
\[ y(t) = 0.1 \sin(10\pi t + \frac{\pi}{2}) \]

### Conclusion
The equation of SHM for the charge `μ` is:
\[ y(t) = 0.1 \sin(10\pi t + \frac{\pi}{2}) \]