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 at a distance `L//4` from one of its ends.
What is the time period of the SHM as mentioned in the above question, if an electric field `E` is applied along horizontal?

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Explanation

To solve the problem, we need to determine the time period of the simple harmonic motion (SHM) of the system described. Here’s a step-by-step solution:

### Step 1: Understand the System
We have an insulating rod of length \( L \) with two small balls of mass \( m \) and charge \( Q \) attached to its ends. The rod can rotate about a vertical axis at a distance \( \frac{L}{4} \) from one end. An electric field \( E \) is applied horizontally.

### Step 2: Identify the Forces Acting on the System
When the rod is displaced from its equilibrium position, the electric force acting on each ball due to the electric field is given by:
\[ F = QE \]
This force will cause a torque about the pivot point.

### Step 3: Calculate the Torque
The torque \( \tau \) due to the electric force on each ball can be calculated. For the ball at a distance \( \frac{3L}{4} \) from the pivot, the torque is:
\[ \tau_1 = \left(\frac{3L}{4}\right) \cdot QE \cdot \sin(\theta) \]
For the ball at a distance \( \frac{L}{4} \) from the pivot, the torque is:
\[ \tau_2 = \left(\frac{L}{4}\right) \cdot QE \cdot \sin(\theta) \]
The net torque \( \tau \) acting on the system is:
\[ \tau = \tau_1 - \tau_2 = \left(\frac{3L}{4}QE - \frac{L}{4}QE\right) \sin(\theta) = \left(\frac{2L}{4}QE\right) \sin(\theta) = \frac{LQE}{2} \sin(\theta) \]

### Step 4: Relate Torque to Angular Acceleration
According to Newton's second law for rotation, the torque is also related to the moment of inertia \( I \) and angular acceleration \( \alpha \):
\[ \tau = I \alpha \]
For the system, the moment of inertia \( I \) can be calculated as:
\[ I = m\left(\frac{L}{4}\right)^2 + m\left(\frac{3L}{4}\right)^2 = m\left(\frac{L^2}{16} + \frac{9L^2}{16}\right) = m\left(\frac{10L^2}{16}\right) = \frac{5mL^2}{8} \]

### Step 5: Set Up the Equation for SHM
Substituting the expression for torque into the equation for angular acceleration gives:
\[ \frac{LQE}{2} \sin(\theta) = \frac{5mL^2}{8} \alpha \]
Since for small angles \( \sin(\theta) \approx \theta \) and \( \alpha = \frac{d^2\theta}{dt^2} \), we can write:
\[ \frac{LQE}{2} \theta = \frac{5mL^2}{8} \frac{d^2\theta}{dt^2} \]

### Step 6: Rearranging to Find the Angular Frequency
Rearranging gives:
\[ \frac{d^2\theta}{dt^2} + \frac{4QE}{5mL} \theta = 0 \]
This is the standard form of SHM, where the angular frequency \( \omega^2 \) is:
\[ \omega^2 = \frac{4QE}{5mL} \]

### Step 7: Calculate the Time Period
The time period \( T \) of SHM is given by:
\[ T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{5mL}{4QE}} \]

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