Past exam of the physics A course of the University of Cambridge / 2019 / Paper 2 / Section B / B7

Here $x(t)$ is displacement from equilibrium, in metres; $m>0$ is the oscillating mass, in kilograms; $b\ge 0$ is the coefficient of linear viscous damping, in $\mathrm{kg}{\mathrm{s}}^{-1}$; and $k>0$ is the spring constant, in $\mathrm{N}{\mathrm{m}}^{-1}$. The terms $m\ddot{x}$, $b\dot{x}$ and $kx$ are respectively the inertial term, the opposing velocity-dependent damping term and the opposing displacement-dependent restoring term when moved to the force side of Newton's second law.

$F_0$ is a possibly complex driving-force complex amplitude, measured in newtons. $\omega$ is the driving angular frequency in $\mathrm{rad}{\mathrm{s}}^{-1}$, and taking the real part of $F_0{e}^{i\omega t}$ gives the physical sinusoidal force. Its frequency in hertz is $f=\omega /(2\pi )$. The natural undamped angular frequency and damping rate areThis damped harmonic oscillator model assumes a linear restoring force and a linear, instantaneous viscous damping law.

Set the drive to zero and try $x=e^{st}$. The characteristic equation isFor underdamping, $\boxed{b^2<4mk}$, the roots are $-\gamma \pm i{\omega }_d$, with ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2}$. The displacement oscillates with an exponentially decaying envelope:For critical damping, $\boxed{b^2=4mk}$, there is a repeated root $-{\omega }_0$, giving $x=(C+Dt)e^{-{\omega }_{0}t}$. The motion is nonoscillatory and is the limiting case between oscillatory and overdamped response.

For overdamping, $\boxed{b^2>4mk}$, the roots are distinct negative real numbers and $x=C{e}^{s_{+}t}+D{e}^{s_{-}t}$. Both components decay without sustained oscillation; strong damping gives a particularly slow late-time relaxation. Constants $C,D$ depend on the initial conditions. A nonoscillatory solution can cross equilibrium once for some initial data, so “nonoscillatory” does not mean monotone for every possible initial condition.

Let the initial conditions be $x(0)=0$ and $\dot{x}(0^{+})=v_0>0$. The velocity-impulse response of a damped oscillator isAll curves start at the origin with the same positive slope $v_0$. Underdamping gives a decaying sequence of positive and negative lobes. Critical damping gives one positive lobe followed by return to zero. Overdamping also remains positive for these initial data, but has a longer tail. The initial impulse changes velocity abruptly, not displacement.

For underdamping, the general response is a free transient plus a particular driven response. Use the paper's $e^{i\omega t}$ convention. The steady displacement complex amplitude isThusHere $\gamma =b/(2m)$, ${\omega }_d=\sqrt{k/m-{\gamma }^2}$ and the real constants $C,D$ enforce the initial conditions. For a real positive $F_0$, the steady term is $|x_0|\cos (\omega t-\varphi )$, whereFor light damping, ${\omega }_d\simeq {\omega }_0$ and the transient decays slowly compared with an oscillation. The two-argument phase convention makes the displacement lag evolve from approximately zero below resonance through $\pi /2$ near ${\omega }_0$ to approximately $\pi$ far above it.

The response is quasistatic at low frequency, resonantly enhanced when the spring and inertia nearly cancel, and suppressed as $1/{\omega }^2$ at high frequency. The original graph gives approximate readings $x_s=|x_0(0)|\simeq 0.10\mathrm{mm}$, $x_p\simeq 1.0\mathrm{mm}$ and $f_r\simeq 150\mathrm{Hz}$. These are graph estimates, not exact supplied measurements.

Use oscillator parameters from static and resonant displacement. The static spring constant isFor light damping, ${\omega }_0\simeq {\omega }_r=2\pi (150)\simeq 9.42\times 10^2{\mathrm{s}}^{-1}$, givingNear resonance, $x_p\simeq F_0/(b{\omega }_0)$, soIndependently, the resonance enhancement $x_p/x_s\simeq 10$ estimates the same quality factor. The half-power displacement level is $x_p/\sqrt{2}\simeq 0.71\mathrm{mm}$; its intersections are approximately $142$ and $157\mathrm{Hz}$, giving bandwidth $\Delta f\simeq 15\mathrm{Hz}$ and $Q\simeq f_0/\Delta f\simeq 10$.

The exact displacement-peak frequency is ${\omega }_r^2={\omega }_0^2-b^2/(2m^2)$. At $Q\simeq 10$ its difference from ${\omega }_0$ is only about $0.25\%$, well below the precision justified by this printed curve. Greater numerical precision for the inferred parameters would be misleading.