There is an insulator rod of length L and of negligible mass with two small balls of mass m and electric charge Q attached to its ends. The rod can rotate in the horizontal plane around a vertical axis crossing it ata distance `L//4` from one of its ends.
What is the time period of the SHM as mentioned in the above question?

Options

A. `2 pi sqrt((mL)/(QE))`
B. `2 pi sqrt((2 mL)/(3 QE))`
C. `2 pi sqrt((5 mL)/(8QE))`
D. `2 pi sqrt((5mL)/(4QE))`

Explanation

To find the time period of the simple harmonic motion (SHM) of the system described, we will follow these steps:

### Step 1: Understand the System
We have an insulator rod of length \( L \) with two small balls of mass \( m \) and charge \( Q \) attached at its ends. The rod can rotate around a vertical axis at a distance \( \frac{L}{4} \) from one end.

### Step 2: Identify the Torque
When the system is displaced by an angle \( \theta \), the torque (\( \tau \)) due to the electric forces acting on the balls can be calculated. The torque due to each ball is given by:
- For the ball at \( \frac{L}{4} \): \( \tau_1 = Q E \cdot \left(\frac{L}{4} \sin \theta\right) \)
- For the ball at \( \frac{3L}{4} \): \( \tau_2 = Q E \cdot \left(\frac{3L}{4} \sin \theta\right) \)

The net torque (\( \tau_{\text{net}} \)) acting on the system is:
\[ \tau_{\text{net}} = \tau_2 - \tau_1 = Q E \cdot \left(\frac{3L}{4} \sin \theta - \frac{L}{4} \sin \theta\right) = Q E \cdot \left(\frac{2L}{4} \sin \theta\right) = \frac{Q E L}{2} \sin \theta \]

### Step 3: Use Small Angle Approximation
For small angles, we can use the approximation \( \sin \theta \approx \theta \):
\[ \tau_{\text{net}} \approx \frac{Q E L}{2} \theta \]

### Step 4: Relate Torque to Angular Acceleration
The torque is also related to angular acceleration (\( \alpha \)) by the equation:
\[ \tau = I \alpha \]
where \( I \) is the moment of inertia of the system.

### Step 5: Calculate the Moment of Inertia
The moment of inertia \( I \) for the two masses about the pivot point is:
\[ I = m \left(\frac{L}{4}\right)^2 + m \left(\frac{3L}{4}\right)^2 = m \left(\frac{L^2}{16} + \frac{9L^2}{16}\right) = m \left(\frac{10L^2}{16}\right) = \frac{5mL^2}{8} \]

### Step 6: Substitute into the Torque Equation
Now substituting the expression for torque into the equation:
\[ \frac{Q E L}{2} \theta = I \alpha \]
We can express \( \alpha \) as \( \frac{d^2\theta}{dt^2} \):
\[ \frac{Q E L}{2} \theta = \frac{5mL^2}{8} \frac{d^2\theta}{dt^2} \]

### Step 7: Rearranging the Equation
Rearranging gives us:
\[ \frac{d^2\theta}{dt^2} + \frac{4Q E}{5mL} \theta = 0 \]
This is the standard form of the SHM equation, where:
\[ \omega^2 = \frac{4Q E}{5mL} \]

### Step 8: Find the Time Period
The time period \( T \) of SHM is given by:
\[ T = 2\pi \sqrt{\frac{1}{\omega^2}} = 2\pi \sqrt{\frac{5mL}{4Q E}} \]

### Final Answer
Thus, the time period of the SHM is:
\[ T = 2\pi \sqrt{\frac{5mL}{4QE}} \]

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