Charged pendulum

Classical Mechanics Level 5

Two point charges, each of charge q and mass m, are hung from 1 m long cords and constrained to move in a 2-d plane (so they can swing but not rotate). The charge and mass are chosen such that the when the charges are in their equilibrium position the angle $\theta_0$ the cords make with the vertical is 0.1 radians. The masses are placed in their equilibrium positions and then each is pulled out by some small angle $\Delta \theta=0.01$ radians and released. What is the period of the resulting oscillation in seconds ?

Details and assumptions

The acceleration of gravity is $-9.8~m/s^2$ .

The answer is 1.42.

1 solution

David Mattingly Staff
May 13, 2014

The total potential energy of the system is given by a combination of the gravitational and electrostatic potential energies, $U=2mgL(1-cos \theta)+kq^2/(2Lsin \theta)$ , where $k=1/(4\pi\epsilon_0$ and L is the length of the cords. Since all the angles are small, we can use the small angle approximation which yields

$U=mgL\theta^2+\frac{kq^2}{2L\theta}$ .

We find the equilibrium position by minimizing U with respect to $\theta$ , which yields

$\theta_0^3=\frac{kq^2}{4mgL^2}$ .

We now can expand the potential energy around this minimum by letting $\theta=\theta_0+\Delta \theta$ . Doing so yields that the potential energy is

$U=U_0+mgL\Delta \theta_0^2$ .

Or, in terms of the horizontal x-displacement,

$U=U_0+\frac {mg}{L}\Delta x^2$ .

However, this is just the potential for an harmonic oscillator. The period of such an oscillator is $T=2\pi\sqrt{\frac{L}{2g}}=1.42~sec$ .

Charged pendulum

The answer is 1.42.

1 solution

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