Charged particle attatched to spring.Find the period of oscillation?

[humanist rho]

Homework Statement

A point particle of mass m carrying an electric charge q is attatched to a spring of stiffness constant k.A constant electric field E aong the direction of spring is switched on for a time interval T (T<<Sqrt(m/k)). Neglecting radiation loss,Find the amplitude of oscillation after the field is switched off.

Homework Equations

The Attempt at a Solution

No idea where to start.Please give me a hint.

Thank you very much.

 

[Curious3141]

Homework Statement

A point particle of mass m carrying an electric charge q is attatched to a spring of stiffness constant k.A constant electric field E aong the direction of spring is switched on for a time interval T (T<<Sqrt(m/k)). Neglecting radiation loss,Find the amplitude of oscillation after the field is switched off.

Homework Equations

The Attempt at a Solution

No idea where to start.Please give me a hint.

Thank you very much.

Treat the direction the electric field is pulling the particle as positive.

Basically, from t = 0 to T, there are two opposing forces acting on the particle, a constant electric force (given by +Eq) and the restoring force of the spring given by -kx. So the net force on the particle is (Eq - kx)

You can now set up the second order d.e. to describe the motion from t = 0 to T :

[itex]m\ddot{x(t)} = Eq - kx(t)[/itex]

Can you construct a solution to this? It's a slight modification of the general equation for SHM.

After this, you can evaluate x(t) at time T to get x(T). After this point, the only force acting on the particle is the restoring force of the spring tugging it "back" to the equilibrium position.

Now figure out how much work is done by the electric force on the particle as it moves from 0 to x(T). Remember that [itex]\int dW = \int Fdx[/itex] and F here is given by Eq, a constant. All this work goes toward increasing the kinetic energy (KE) of the particle.

To determine the position [itex]x_0[/itex] at which the particle comes to rest (this is the maximal amplitude of the subsequent oscillation), remember the principle of conservation of energy. At the position [itex]x_0[/itex], the kinetic energy imparted by the electrical force has been completely converted to stored potential energy in the spring. Use PE = [itex]\frac{1}{2}k{x_0}^2[/itex] to express the latter quantity.

Now you can relate the KE to the PE. You end up with a quadratic equation in [itex]x_0[/itex], which is the maximal amplitude you want to determine.

After this point, of course, the particle just continues to oscillate with that amplitude and at the system's natural frequency (given by [itex]\omega = \sqrt{\frac{k}{m}}[/itex], or equivalently, [itex]f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}[/itex] and [itex]T = 2\pi\sqrt{\frac{m}{k}}[/itex]).

The importance of stating that [itex]T << \sqrt{\frac{m}{k}}[/itex] (i.e. time that electrical field is ON is much less than the natural period of the oscillator) is to ensure that the electrical force doesn't continue to interfere with the natural motion of the spring system for "too long", otherwise the oscillations will not be SHM, and the amplitude calculation will be off.