Past exam of the physics A course of the University of Cambridge / 2018 / Paper 2 / Section B / B6

Substituting $x=A{e}^{pt}$ into the damped harmonic oscillator equation givesFor nonnegative viscous damping, the cases areThe PDF's request to express these conditions in terms of $\gamma /m$ is dimensionally inconsistent with its definition $\gamma =b/m$: $\gamma$ already has units ${\mathrm{s}}^{-1}$. With the unnormalised equation $m\ddot{x}+b\dot{x}+kx=0$, the equivalent comparisons are $b/m\gtrless 2{\omega }_0$, or $b^2\gtrless 4mk$. Also ${\omega }_0$ is the undamped natural angular frequency, not in general the driven displacement peak.

For the steady response of a forced harmonic oscillator, write $x=\mathrm{Re}(A{e}^{i\omega t})$ and use the same complex amplitude convention for the driving force. Substitution givesThus the response magnitude and displacement phase relative to the force areThe phase goes from $0$ at low frequency, through $-\pi /2$ at ${\omega }_0$, towards $-\pi$ at high frequency. The static magnitude is $1/(m{\omega }_0^2)=1/k$, while the high-frequency magnitude is approximately $1/(m{\omega }^2)$.

Define the quality factor by $Q={\omega }_0/\gamma$. If $Q>1/\sqrt{2}$, the displacement peak occurs at ${\omega }_r=\sqrt{{\omega }_0^2-{\gamma }^2/2}$; for $Q\le 1/\sqrt{2}$ the magnitude decreases monotonically from its static value. For $Q\gg 1$ there is a tall narrow peak close to ${\omega }_0$, of height approximately $Q/k$ and width approximately $\gamma$. At very small $Q$ the resonant peak disappears and the response is broad. The plotted functions are normalized as $k|R|$, at fixed $m,k$.

The original diagram is a series RLC circuit, so the same electric current $i=\dot{q}$ passes through all three elements. Choose capacitor electric charge $q$ positive on the plate into which positive $i$ enters. Kirchhoff's voltage law givesDividing by inductance $L$ givesThe correspondence to the forced harmonic oscillator is $m\leftrightarrow L$, $b\leftrightarrow R$, $k\leftrightarrow 1/C$, and $x\leftrightarrow q$. The stored energy is $L{i}^2/2+q^2/(2C)$; its balance is $\dot{E}=Vi-R{i}^2$, confirming the damping sign.

The capacitor-charge complex amplitude isFor fixed driving-voltage amplitude, maximizing $|q_0|$ minimizesDifferentiate:The positive-frequency minimum therefore hasThis is the resonance of charge, or capacitor voltage. If the stated inequality fails, their maximum is at zero frequency, and no positive displacement-resonance peak exists. The electric current has the additional factor $\omega$, and instead peaks exactly at ${\omega }_0=1/\sqrt{LC}$ for every $R>0$.

The precise bandwidth depends on the output: this is the distinction between displacement and current bandwidths of a series RLC circuit. For current output, or voltage across the resistor, the electrical impedance givesThe half-power points satisfy $|\omega L-1/(\omega C)|=R$. Their positive roots areChanging capacitance shifts the centre without changing this absolute width. For the charge response in part (d), the exact half-height points instead obeywhen both exist their separation only approaches $\gamma =R/L$ at large quality factor. Therefore the printed capacitance-independence claim is exact for current or resistor-voltage output and an approximation for the preceding charge resonance.

For a narrow band-pass filter, use current or resistor-voltage output and make $R/L$ small, counting source, load and inductor losses in the total electrical resistance. Increase $Q=\sqrt{L/C}/R$; if a specified centre frequency is to remain fixed, adjust $C$ with $L$ to keep $LC$ fixed. Capacitor-voltage output alone is not a band-pass response because it has a nonzero DC limit.