Oscillations Horizontal Spring Capacitor

Horizontal Spring–Mass System attached to one of the Capacitor plates

Initial Position

The spring is unstretched.$$x=0$$.Capacitor plate carries charge \(+Q\).The opposite plate carries charge \(-Q\). Electrostatic attraction pulls the movable plate toward the fixed plate.

\[C=\frac{\varepsilon_0A}{d}\] \[U_i=\frac12\frac{Q^2}{C}\] \[U_i=\frac12\frac{Q^2}{\varepsilon_0 A}\left(x_{extreme}+d\right)\]

Mean Position

At the mean position the net force becomes zero.

\[F_{spring}=F_{attraction}\] \[kx_{mean}=\frac12\frac{Q^2}{\varepsilon_0A}\] \[\boxed{x_{mean}=\frac{Q^2}{2\varepsilon_0Ak}}\]

This is the equilibrium position about which oscillation occurs.

Extreme Position

At the extreme position the velocity becomes zero.Difference in Capacitor energy is stored as spring potential energy.

\[PE_{spring}+KE=U_i-U_f\] \[\frac12 kx_{extreme}^{\,2}+0=\frac12\frac{Q^2}{\varepsilon_0A}\left(x_{extreme}+d\right) -\frac12\frac{Q^2}{\varepsilon_0A}d\] \[\frac12 kx_{extreme}^{\,2}=\frac12\frac{Q^2}{\varepsilon_0A}x_{extreme} \] \[\boxed{x_{extreme}=\frac{Q^2}{\varepsilon_0Ak}}\] \[x_{extreme}=2 \, x_{mean}\]

\(F_{\text{attraction}}\) between Capacitor plates

\[ U = \frac{1}{2}\frac{Q^2}{C} \\[0.7em] \] \[ W = \frac{1}{2}\frac{Q^2}{\varepsilon_0 A}d \\[0.7em] \] \[ Fd = \frac{1}{2}\frac{Q^2}{\varepsilon_0 A}d \\[0.7em] \] \[ \boxed{F_{\text{attraction}} = \frac{1}{2}\frac{Q^2}{\varepsilon_0 A} }\\[0.7em] \]

Max Velocity occurs at Mean Position

\[ P.E._{\text{spring}} + K.E._{\text{mean}} = U_i - U_f \\[0.7em] \] \[ \frac{1}{2}kx_{\text{mean}}^2 + \frac{1}{2}mv_{\text{mean}}^2 = \frac{1}{2}\frac{Q^2}{\varepsilon_0A}(x_{\text{extreme}}+d) -\frac{1}{2}\frac{Q^2}{\varepsilon_0A}(x_{\text{mean}}+d) \\[0.7em] \] \[ \frac{1}{2}kx_{\text{mean}}^2 + \frac{1}{2}mv_{\text{mean}}^2 = \frac{1}{2}\frac{Q^2}{\varepsilon_0A} \left(x_{\text{extreme}}-x_{\text{mean}}\right) \\[0.7em] \] \[ \frac{1}{2}kx_{\text{mean}}^2 + \frac{1}{2}mv_{\text{mean}}^2 = \frac{1}{2}\frac{Q^2}{\varepsilon_0A}\,x_{\text{mean}} \\[0.7em] \] \[ k\left(\frac{Q^2}{2\varepsilon_0Ak}\right)^2 + mv_{\text{mean}}^2 = \frac{Q^2}{\varepsilon_0A} \left(\frac{Q^2}{2\varepsilon_0Ak}\right) \\[0.7em] \] \[ mv_{\text{mean}}^2 = \frac{Q^4}{2\varepsilon_0^2A^2k} - \frac{Q^4}{4\varepsilon_0^2A^2k} \\[0.7em] \] \[ mv_{\text{mean}}^2 = \frac{Q^4}{4\varepsilon_0^2A^2k} \\[0.7em] \] \[ \boxed{v_{\text{mean}} = \frac{Q^2}{2\varepsilon_0A} \sqrt{\frac{1}{km}} \frac{1}{km}}\\[0.7em] \]

Key Idea:

Important Observations

Increasing charge \(Q\) increases both mean position and amplitude. Increasing spring constant \(k\) reduces displacement. A stiffer spring produces smaller oscillations. The equilibrium position lies exactly halfway between the natural length position and the extreme position. The oscillation occurs about the mean position.