Question

A simple pendulum of mass m and charge +q is suspended vertically by a massless thread of length l. At the point of suspension, a point charge +q is also fixed. If the pendulum is displaced slightly from equilibrium position, its time period will be

Answer 1

The time period is  T = 2 π √ l / g

Explanation:

• Mass of pendulum = m
• Charge on pendulum = +q
• Length of pendulum = l
• To Find: Time period "T" = ?

Solution:

The electrostatic force is along the radial direction. But the oscillation movement is along the tangential direction.

Now the work done by the electrostatic force is

Wc = F . d = fd cos 90°

Since cos 90° = 0

Therefore Wc = 0

The work done is always along the radial direction.

The time period is as like the time period of normal pendulum i.e.

T = 2 π √ l / g