Spring mass in Electric field Total Energy

 

Amplitude is maximum displacement from equilibrium

\(F_s=-kx\)

\(ma=-kx\)

\(a=-\frac{k}{m}x\)

\(ma=-(kx-qE)\)

\(a=\frac{qE}{m}-\frac{k}{m}x\)

Shifting from initial position

At equilibrium, \(a=0\)

\(x_{eq}=\frac{qE}{k}\)

\(a=\frac{dv}{dt}\)

\(\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}\)

\(v\frac{dv}{dx}=\left(\frac{qE}{m}-\frac{kx}{m}\right)\)

\(\int v\,dv=\int\left(\frac{qE}{m}-\frac{kx}{m}\right)dx\)

\(\frac{v^2}{2}=\frac{qE}{m}x-\frac{k}{2m}x^2+C\)

To find constant \(C\) (non-general constant)

At extreme position, \(v=0\), when \(x=x_{eq}+A\)

\(0=\frac{qE}{m}x-\frac{k}{2m}x^2+C\)

\(2qEx-kx^2+2Cm=0\)

\(2qE\left(\frac{qE}{k}+A\right)-k\left(\frac{qE}{k}+A\right)^2+2Cm=0\)

\(\frac{2q^2E^2}{k}+2qEA-\left(\frac{q^2E^2}{k}+kA^2+2qEA\right)+2Cm=0\)

\(\frac{q^2E^2}{k}-kA^2+2Cm=0\)

\(C=\frac{kA^2}{2m}-\frac{q^2E^2}{2mk}\)

\(PE_{equil}=-\left(F_sx_{eq}+F_ex_{eq}\right)\)

\(=\frac{1}{2}kx_{eq}^2-qEx_{eq}\)

\(=\frac{1}{2}k\frac{q^2E^2}{k^2}-qE\frac{qE}{k}\)

\(=\frac{1}{2}\frac{q^2E^2}{k}-\frac{q^2E^2}{k}\)

\(PE_{equil}=-\frac{1}{2}\frac{q^2E^2}{k}\)

\(PE_{total}=PE_{spring}+PE_{equil}\)

\(=\frac{1}{2}kA^2-\frac{1}{2}\frac{q^2E^2}{k}\)

\(PE_{total}=\frac{1}{2}m\omega^2A^2-\frac{1}{2}\frac{q^2E^2}{k}\)

\(PE_{extreme}=PE_{spring}+PE_{electric}\)

For \(x=x_{eq}-A\)

\(PE_{extreme}=\frac{1}{2}kx^2-qEx\)

\(=\frac{1}{2}k\left(\frac{qE}{k}-A\right)^2-qE\left(\frac{qE}{k}-A\right)\)

\(=\frac{1}{2}\frac{q^2E^2}{k}+\frac{1}{2}kA^2-qEA-\frac{q^2E^2}{k}+qEA\)

\(=\frac{1}{2}kA^2-\frac{1}{2}\frac{q^2E^2}{k}\)

\(PE_{extreme}=\frac{1}{2}m\omega^2A^2-\frac{1}{2}\frac{q^2E^2}{k}\)

For \(v=v_{max}\)

\(\int_{0}^{v_{max}}v\,dv=\int_{x_{eq}}^{x_{eq}+A}\left(\frac{qE}{m}-\frac{kx}{m}\right)dx\)

\(\left[\frac{v^2}{2}\right]_{0}^{v_{max}}=\left[\frac{qE}{m}x-\frac{kx^2}{2m}\right]_{x_{eq}}^{x_{eq}+A}\)

\(\frac{1}{2}v_{max}^2=\left[\frac{qE}{m}(x_{eq}+A)-\frac{k}{2m}(x_{eq}+A)^2\right]-\left[\frac{qE}{m}x_{eq}-\frac{k}{2m}x_{eq}^2\right]\)

\(=\frac{qEA}{m}-\frac{kA^2}{2m}\)

\(-\left[\frac{q^2E^2}{km}+\frac{qEA}{m}-\frac{q^2E^2}{2km}\right]\)

\(\frac{1}{2}v_{max}^2=\frac{kA^2}{2m}\)

\(K.E_{max}=\frac{1}{2}mv_{max}^2=\frac{1}{2}kA^2\)

\(K.E_{equil}=\frac{1}{2}m\omega^2A^2\)

\(V_{max}=A\omega=A\sqrt{\frac{k}{m}}\)

Comparing with previous method finding