Exercises — LC circuit — oscillations (electrical analog of SHM)

WHAT we compare: mechanical $m\ddot{x}+kx=0$ vs electrical $L\ddot{q}+\frac{1}{C}q=0$. Line them up term by term: the coefficient in front of the second derivative is the "inertia," the coefficient in front of the variable itself is the "stiffness."

• Inertia (mass): $m\leftrightarrow ==L==$ (the inductor resists change in current, exactly as mass resists change in velocity).
• Stiffness (spring): $k\leftrightarrow ==1/C==$ (a small $C$ is a stiff spring — it fights hard to change its voltage).

Formula: $\omega =\frac{1}{\sqrt{LC}}$. Why this one? It is the direct read-off from matching $\ddot{q}=-\frac{1}{LC}q$ to $\ddot{x}=-{\omega }^{2}x$. $LC=(4\times 10^{-3})(1\times 10^{-6})=4\times 10^{-9}{\text{s}}^2$ $\sqrt{LC}=\sqrt{4\times 10^{-9}}=2\times 10^{-4.5}=6.32\times 10^{-5}\text{s}$ $\omega =\frac{1}{6.32\times 10^{-5}}\approx 1.58\times 10^4\text{rad/s}$

False. Current $i=\dot{q}$ is the slope of $q(t)=q_0\cos \omega t$; at the peak of a cosine the slope is zero, so $i=0$ there. Current peaks when $q=0$ (where the cosine crosses zero and is steepest). They are $90^{\circ }$ apart.

$LC=(2\times 10^{-3})(8\times 10^{-6})=1.6\times 10^{-8}{\text{s}}^2,\sqrt{LC}=1.2649\times 10^{-4}\text{s}$ $T=2\pi \sqrt{LC}=2\pi (1.2649\times 10^{-4})\approx 7.95\times 10^{-4}\text{s}(\approx 0.795\text{ms})$ $f=\frac{1}{T}\approx 1258\text{Hz}$ Why $T=2\pi \sqrt{LC}$ and not $2\pi /\sqrt{LC}$? Because $T=2\pi /\omega$ and $\omega =1/\sqrt{LC}$; dividing by $1/\sqrt{LC}$ multiplies by $\sqrt{LC}$, so the root lands in the numerator.

$\omega =\frac{1}{\sqrt{LC}}=\frac{1}{1.2649\times 10^{-4}}=7905.7\text{rad/s}$ $i_0=q_0\omega =(4\times 10^{-4})(7905.7)\approx 3.16\text{A}$ Why $i_0=q_0\omega$? Differentiate $q=q_0\cos \omega t$: the amplitude of $-q_0\omega \sin \omega t$ is $q_0\omega$. Equivalently, all electric energy $q_0^2/2C$ becomes magnetic energy $\frac{1}{2}L{i}_0^2$ at the crossing.

With $q_0=4\times 10^{-4}$ C, $\omega =7906$ rad/s, $i_0=3.16$ A: $q(t)=(4\times 10^{-4})\cos (7906t)\text{C}$ $i(t)=-(3.16)\sin (7906t)\text{A}$ The minus sign says the current first flows in the direction that discharges the capacitor.

WHAT we want: $U_E=U_B$. Write both: $U_E=\frac{q_0^2}{2C}{\cos }^2\omega t,U_B=\frac{q_0^2}{2C}{\sin }^2\omega t$ WHY they take this form: using $L{\omega }^2=1/C$ turns $\frac{1}{2}L{i}^2$ into the same prefactor $q_0^2/2C$ times ${\sin }^2$. Set them equal: ${\cos }^2\omega t={\sin }^2\omega t\Rightarrow {\tan }^2\omega t=1\Rightarrow \tan \omega t=1$ (first positive solution) $\Rightarrow \omega t=\frac{\pi }{4}$. $t=\frac{\pi }{4\omega }=\frac{\pi }{4\times 10^4}\approx 7.85\times 10^{-5}\text{s}=\frac{T}{8}$ Look at the figure: the two curves cross first at $T/8$, where each holds exactly half the total energy.

WHY energy, not time: we are asked "what $q$ when $i=i_0/2$," a relation between $q$ and $i$ — energy links them directly. $\underset{U_E}{\underbrace{\frac{q^2}{2C}}}+\underset{U_B}{\underbrace{\frac{1}{2}L{i}^2}}=\frac{q_0^2}{2C}$ At $i=i_0/2$, and using $\frac{1}{2}L{i}_0^2=\frac{q_0^2}{2C}$ so $\frac{1}{2}L(i_0/2{)}^2=\frac{1}{4}\cdot \frac{q_0^2}{2C}$: $\frac{q^2}{2C}=\frac{q_0^2}{2C}-\frac{1}{4}\frac{q_0^2}{2C}=\frac{3}{4}\frac{q_0^2}{2C}\Rightarrow q=\frac{\sqrt{3}}{2}{q}_0$ Numerically $q=\frac{\sqrt{3}}{2}(2\times 10^{-4})\approx 1.73\times 10^{-4}$ C.

$\omega =\frac{1}{\sqrt{LC}}$ depends only on the product $LC$. To keep $\omega$ fixed, keep $LC$ fixed. If $L\to 2L$, then $C\to C/2$ (halve it), because $(2L)(C/2)=LC$. Halve the capacitance.

Step 1 (WHAT): Go once around the closed loop; by Kirchhoff's Voltage Law the voltage rises and drops must sum to zero: $V_C+V_L=0\Rightarrow \frac{q}{C}+L\frac{di}{dt}=0$ WHY: with no battery and no resistor, the only two voltages are the capacitor's ($q/C$) and the inductor's back-EMF ($Ldi/dt$); they must cancel. Step 2 (WHAT): Replace $i=\frac{dq}{dt}$, so $\frac{di}{dt}=\frac{d^{2}q}{d{t}^2}$: $L\frac{d^{2}q}{d{t}^2}+\frac{q}{C}=0\Rightarrow \frac{d^{2}q}{d{t}^2}=-\frac{1}{LC}q$ WHY: current is the rate of charge change, so its rate of change is the second derivative of $q$. Step 3 (WHAT): Compare with the SHM template $\ddot{x}=-{\omega }^{2}x$ from Simple Harmonic Motion. Matching coefficients, ${\omega }^2=\frac{1}{LC}$, so $\omega =\frac{1}{\sqrt{LC}}$.

Find $C$ from the period. $T=2\pi \sqrt{LC}\Rightarrow \sqrt{LC}=\frac{T}{2\pi }\Rightarrow LC={\left(\frac{T}{2\pi }\right)}^2$. $\frac{T}{2\pi }=\frac{4\times 10^{-4}}{2\pi }=6.366\times 10^{-5}\text{s},LC=4.053\times 10^{-9}{\text{s}}^2$ $C=\frac{LC}{L}=\frac{4.053\times 10^{-9}}{1\times 10^{-3}}=4.053\times 10^{-6}\text{F}\approx 4.05\mu \text{F}$ Find $i_0$. First $q_0=C{V}_0=(4.053\times 10^{-6})(20)=8.11\times 10^{-5}$ C. Then $\omega =\frac{2\pi }{T}=\frac{2\pi }{4\times 10^{-4}}=1.5708\times 10^4\text{rad/s}$ $i_0=q_0\omega =(8.11\times 10^{-5})(1.5708\times 10^4)\approx 1.27\text{A}$

$U_E=\frac{q_0^2}{2C}{\cos }^2\omega t,U_B=\frac{1}{2}L(q_0\omega {)}^2{\sin }^2\omega t=\frac{L{\omega }^2{q}_0^2}{2}{\sin }^2\omega t$ Use $L{\omega }^2=L\cdot \frac{1}{LC}=\frac{1}{C}$, so $\frac{L{\omega }^2{q}_0^2}{2}=\frac{q_0^2}{2C}$. Then $U_E+U_B=\frac{q_0^2}{2C}({\cos }^2\omega t+{\sin }^2\omega t)=\frac{q_0^2}{2C}$ since ${\cos }^2+{\sin }^2=1$. The constant total energy is $\boxed{U=\frac{q_0^2}{2C}}$.

WHY resonance: $f=\frac{1}{2\pi \sqrt{LC}}$ is the natural frequency — this is the resonant frequency the circuit rings at. Solve for $C$: square and rearrange. $f=\frac{1}{2\pi \sqrt{LC}}\Rightarrow \sqrt{LC}=\frac{1}{2\pi f}\Rightarrow LC=\frac{1}{4{\pi }^2{f}^2}\Rightarrow C=\frac{1}{4{\pi }^2{f}^{2}L}$ $C=\frac{1}{4{\pi }^2(1.0\times 10^6{)}^2(25\times 10^{-6})}=\frac{1}{4{\pi }^2\cdot 10^{12}\cdot 25\times 10^{-6}}$ $=\frac{1}{9.8696\times 10^8}\approx 1.013\times 10^{-9}\text{F}\approx 1.01\text{nF}$

(a) $C\to \infty$ (huge capacitor = super-soft spring). $\omega =1/\sqrt{LC}\to 0$: the period $T=2\pi \sqrt{LC}\to \infty$. The "spring" $1/C\to 0$ vanishes, so there is nothing to pull the charge back — oscillation becomes infinitely slow. (b) $L\to 0$ (no inductor = no inertia). $\omega \to \infty$, $T\to 0$. With no electrical inertia there is nothing to make the current overshoot; the capacitor would just discharge instantly — the oscillation degenerates. (c) Add resistance $R$. Energy is no longer conserved; the circuit becomes an LCR Circuit showing Damped Oscillations. The amplitude decays and the frequency shifts slightly to ${\omega }_d=\sqrt{\frac{1}{LC}-\frac{R^2}{4L^2}}$. The ideal LC result is the $R\to 0$ limit of this.

(i) Total energy. $U=\frac{q_0^2}{2C}=\frac{(2\times 10^{-4}{)}^2}{2(5\times 10^{-6})}=\frac{4\times 10^{-8}}{1\times 10^{-5}}=4\times 10^{-3}\text{J}=4\text{mJ}$ (ii) Quarter energy in the capacitor. Need $U_E=\frac{1}{4}U$: $\frac{q_0^2}{2C}{\cos }^2\omega t=\frac{1}{4}\frac{q_0^2}{2C}\Rightarrow {\cos }^2\omega t=\frac{1}{4}\Rightarrow \cos \omega t=\frac{1}{2}$ (take the positive root for the first crossing) $\Rightarrow \omega t=\frac{\pi }{3}$. $t=\frac{\pi }{3\omega }=\frac{\pi }{3\times 10^4}\approx 1.047\times 10^{-4}\text{s}$ As a fraction of $T$: $\frac{t}{T}=\frac{\omega t}{2\pi }=\frac{\pi /3}{2\pi }=\frac{1}{6}$. So it happens at $\boxed{T/6}$.