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`.
Frequency of oscillation is

Options

A. 8
B. 10
C. 5
D. 2

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant physics concepts.

### Step 1: Understand the System
We have two equal point charges, each \( Q = 8 \, \mu C = 8 \times 10^{-6} \, C \), located at points \( (2, 0) \) and \( (-2, 0) \) on the x-axis. There is another charge \( \mu \) with a mass of \( 91 \, mg = 91 \times 10^{-6} \, kg \) positioned at \( (0, 0.1) \) on the y-axis.

### Step 2: Calculate the Force on Charge \( \mu \)
The force experienced by charge \( \mu \) at \( (0, 0.1) \) is given as \( F = 9 \times 10^{-3} \, N \).

### Step 3: Determine the Acceleration
Using Newton's second law, we can find the acceleration \( a \) of charge \( \mu \):
\[ a = \frac{F}{m} \]
Substituting the values:
\[ a = \frac{9 \times 10^{-3} \, N}{91 \times 10^{-6} \, kg} = \frac{9 \times 10^{-3}}{91 \times 10^{-6}} \approx 98.9 \, m/s^2 \]

### Step 4: Identify the Displacement
The displacement from the mean position (the equilibrium position on the y-axis) is given as \( y = 0.1 \, m \).

### Step 5: Calculate the Time Period of Oscillation
For simple harmonic motion, the time period \( T \) can be calculated using the formula:
\[ T = 2\pi \sqrt{\frac{y}{a}} \]
Substituting the values of displacement and acceleration:
\[ T = 2\pi \sqrt{\frac{0.1}{98.9}} \]
Calculating the square root:
\[ \sqrt{\frac{0.1}{98.9}} \approx \sqrt{0.00101} \approx 0.0318 \]
Now substituting back into the time period formula:
\[ T \approx 2\pi \times 0.0318 \approx 0.199 \]

### Step 6: Calculate the Frequency of Oscillation
The frequency \( f \) is the reciprocal of the time period:
\[ f = \frac{1}{T} \approx \frac{1}{0.199} \approx 5.03 \, Hz \]
Rounding off, we can say:
\[ f \approx 5 \, Hz \]

### Final Answer
The frequency of oscillation is approximately \( 5 \, Hz \).

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