= Solution
The restoring, <damping> and <electric forces> on the <electron> give $m_e\ddot x=-kx-b\dot x-eE(t)$. Dividing by $m_e$ gives a <Lorentz oscillator model> with
$$
\omega_0=\sqrt{k/m_e}=\boxed{1.785\times10^{15}\,\mathrm{rad\,s^{-1}}},\qquad
\gamma=b/m_e=\boxed{6.264\times10^{12}\,\mathrm{s^{-1}}}.
$$
For the $e^{i\omega t}$ convention, substitute $x=\operatorname{Re}(x_0e^{i\omega t})$. Then $x_0=-eE_0/[m_e(\omega_0^2-\omega^2+i\gamma\omega)]$ and
$$
\boxed{v_0=i\omega x_0=\frac{-i\omega eE_0}{m_e(\omega_0^2-\omega^2+i\gamma\omega)}.}
$$
The time-averaged dissipated <power> is
$$
P=\frac{be^2|E_0|^2}{2m_e^2}\frac{\omega^2}{(\omega_0^2-\omega^2)^2+\gamma^2\omega^2},\qquad
\frac P{P_{\max}}=\frac{\gamma^2}{(\omega_0^2/\omega-\omega)^2+\gamma^2}.
$$
For positive <frequency> the squared term is minimized to zero only at \b[$\omega=\omega_0$]. Thus the <power> maximum is exactly at the undamped <natural frequency>, even though the <displacement> maximum is slightly shifted. Here $P_{\max}=e^2|E_0|^2/(2b)$. The half-power equation is $|\omega_0^2-\omega^2|=\gamma\omega$, whose two positive roots are
$$
\omega_\pm=\frac{\sqrt{4\omega_0^2+\gamma^2}\pm\gamma}{2},\qquad
\boxed{\Delta\omega=\omega_+-\omega_-=\gamma.}
$$
This is the <absorbed-power bandwidth of a viscously damped oscillator>, exact within the model. Near <resonance> the profile is approximately a <Lorentzian line shape>. With $c=2.99792458\times10^8\,\mathrm{m\,s^{-1}}$,
$$
\boxed{\lambda_{\rm peak}=\frac{2\pi c}{\omega_0}=1055.2\,\mathrm{nm}},\qquad
\boxed{\Delta\lambda=2\pi c\left(\frac1{\omega_-}-\frac1{\omega_+}\right)=\frac{2\pi c\gamma}{\omega_0^2}=3.70\,\mathrm{nm}.}
$$
For this <wavelength linewidth of a damped-oscillator power resonance> we used $\omega_+\omega_-=\omega_0^2$. The half-power <wavelength> endpoints are not exactly symmetric around $\lambda_{\rm peak}$, although that distinction is tiny here.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2011/paper-1-oscillator.png]
{title=Absorption resonance and release of a polarized oscillator}
{description=Original model curves. Frequency is offset from resonance in linewidth units. The release panel uses the supplied mechanical parameters and a unit initial displacement.}
In a constant field the settled <displacement> is $x_s=-eE_{\rm dc}/k$. On sudden removal, with zero initial <velocity>, the free motion is
$$
x(t)=x_s e^{-\gamma t/2}\left[\cos\Omega t+\frac{\gamma}{2\Omega}\sin\Omega t\right],\qquad
\Omega=\sqrt{\omega_0^2-\gamma^2/4}.
$$
The <electron> oscillates about its original equilibrium, losing the previously stored spring <energy>. The period is \b[$3.52\,\mathrm{fs}$], the amplitude decay time is \b[$2/\gamma=0.319\,\mathrm{ps}$], and the cycle-averaged <energy> decay time is \b[$1/\gamma=0.160\,\mathrm{ps}$]. This is the effective classical model supplied, not a prediction of the full hydrogen line <radiation spectrum>.
Back to article page
