Electric Potential Energy — Question 9

Question 9

A uniform field shifts a spring equilibrium and releases mechanical energy

A particle of mass m and charge q is attached to a spring with constant $\kappa$ along x. A constant uniform field E points along positive x. Take $U(x)=\kappa x^2/2-qEx$. Release the particle from rest at x=0.

1. Complete the square to find the equilibrium and minimum energy.

2. Find the oscillation amplitude about equilibrium and the maximum speed.

Question 9 - Solution

A uniform field shifts a spring equilibrium and releases mechanical energy

(a) Method and calculation

$$U=\frac\kappa2\left(x-\frac{qE}{\kappa}\right)^2-\frac{(qE)^2}{2\kappa}.$$$$x_{\mathrm{eq}}=qE/\kappa,\quad U_{\min}=-(qE)^2/(2\kappa).$$

(b) Interpretation and further calculation

$$A=|x_{\mathrm{eq}}|,\quad K_{\max}=U(0)-U_{\min}=\frac{(qE)^2}{2\kappa}.$$$$v_{\max}=\frac{|qE|}{\sqrt{m\kappa}},\quad\omega=\sqrt{\kappa/m}.$$

Physical check

The field changes the equilibrium and stored energy but not the curvature of the spring energy or its oscillation frequency.