A spring and an electric circuit seem like different problems. But displacement in one and charge in the other obey the same kind of second-order equation. The characteristic-root and particular-solution methods let us predict whether they oscillate, settle down, or keep responding to an outside input.

Building the mass–spring model

Hang a mass $m>0$ from a vertical spring. Let $l$ be the spring's natural length and $L$ its extra stretch once the mass is hanging at rest. Measure the moving mass's displacement $y(t)$ from that equilibrium position, with downward positive.

These are three different lengths: the natural length is $l$, the equilibrium length is $l+L$, and the moving spring's length is $l+L+y$. The total stretch relative to the natural length is therefore $L+y$, not just $y$.

At rest, the upward spring force balances the weight. With gravitational acceleration $g$ and spring constant $k>0$, Hooke's law gives

$$mg-kL=0, \qquad \boxed{k=\frac{mg}{L}}.$$

That lets us find the stiffness from a measured static stretch. In SI units, $m$ is in kilograms, $L$ in meters, and $k$ in newtons per meter.

Forces during motion

Once the mass moves, include all four forces with the same sign convention:

The damping coefficient $\gamma\ge0$, read “gamma,” has units $\mathrm{N\,s/m}$. The value $\gamma=0$ represents the undamped idealization. Notice that $y<0$ does not necessarily mean the spring is compressed: it is still stretched if $L+y>0$.

Newton's second law says mass times acceleration equals net force:

$$\begin{aligned} my''&=mg-k(L+y)-\gamma y'+F(t)\\ &=(mg-kL)-ky-\gamma y'+F(t). \end{aligned}$$

The static balance makes $mg-kL=0$. Rearranging gives

$$\boxed{my''+\gamma y'+ky=F(t)}.$$

Gravity has not stopped acting. Measuring displacement from equilibrium has already accounted for its constant contribution.

To determine a motion, specify initial displacement and velocity:

$$y(0)=y_0,\qquad y'(0)=v_0.$$

For example, pulling the mass downward gives $y_0>0$, an upward initial velocity gives $v_0<0$, and “released from rest” means $v_0=0$. With constant $m>0$, $k>0$, $\gamma\ge0$ and continuous $F$ on an interval containing $0$, the linear existence-and-uniqueness theorem gives exactly one solution on that interval.

This is a model, not an exact description of every spring. It neglects the spring's mass and assumes a linear force–stretch relation and velocity-proportional drag. Those approximations are useful within the relevant small-deformation and low-speed regimes.

Example: turning measurements into an IVP

A $2\,\mathrm{kg}$ mass stretches a spring by $4.9\,\mathrm{cm}$. A dashpot exerts $8\,\mathrm N$ of resistance at a speed of $0.5\,\mathrm{m/s}$. The mass is pulled $5\,\mathrm{cm}$ below equilibrium and released, with no external forcing. Use $g=9.8\,\mathrm{m/s^2}$.

Convert the lengths to meters before computing:

$$k=\frac{2(9.8)}{0.049}=400\,\mathrm{N/m}, \qquad \gamma=\frac{8}{0.5}=16\,\mathrm{N\,s/m}.$$

The dashpot's calibration speed is not the initial velocity. The mass is released from rest, so the IVP is

$$2y''+16y'+400y=0, \qquad y(0)=0.05,\qquad y'(0)=0.$$

Dividing by the mass gives the equivalent normalized equation

$$y''+8y'+200y=0.$$

The physical damping coefficient is still $\gamma=16$; $8=\gamma/m$ is the normalized coefficient. We will solve this IVP after distinguishing the damping cases.

Free motion without damping

Free means $F(t)=0$: no applied forcing beyond the forces already built into the model. Undamped means $\gamma=0$. Together these give

$$my''+ky=0, \qquad mr^2+k=0.$$

The characteristic roots are $r=\pm i\sqrt{k/m}$. Define the natural angular frequency

$$\omega_0=\sqrt{\frac{k}{m}}.$$

The solution is simple harmonic motion:

$$\boxed{y(t)=C_1\cos(\omega_0t)+C_2\sin(\omega_0t)}.$$

For a nonzero motion, the amplitude is $\sqrt{C_1^2+C_2^2}$ and the period is

$$\boxed{T=\frac{2\pi}{\omega_0}=2\pi\sqrt{\frac{m}{k}}}.$$

Angular frequency is in radians per second; frequency in cycles per second is $1/T=\omega_0/(2\pi)$. A heavier mass oscillates more slowly, while a stiffer spring oscillates faster. Initial conditions set amplitude and phase, not the natural frequency.

Example: an upward initial velocity

Let $m=0.5\,\mathrm{kg}$ and $k=8\,\mathrm{N/m}$. Start $0.1\,\mathrm m$ below equilibrium, moving upward at $0.4\,\mathrm{m/s}$. Then

$$\frac12y''+8y=0, \qquad y(0)=0.1,\qquad y'(0)=-0.4.$$

Here $\omega_0=4\,\mathrm{rad/s}$. Write $y=C_1\cos4t+C_2\sin4t$. The initial conditions give

$$C_1=0.1,\qquad 4C_2=-0.4,$$

so

$$\boxed{y(t)=0.1\cos4t-0.1\sin4t\quad\mathrm m}.$$

The period is $\pi/2\,\mathrm s\approx1.57\,\mathrm s$, and the amplitude is $0.1\sqrt2\,\mathrm m$. It is larger than the initial displacement because the mass also starts with kinetic energy.

Energy and damping

For displacement measured from equilibrium, the mechanical energy, up to an irrelevant constant, is

$$\mathcal E(t)=\underbrace{\frac12m(y')^2}_{\text{kinetic}} +\underbrace{\frac12ky^2}_{\text{potential}}.$$

To see how it changes, multiply the equation of motion by velocity $y'$:

$$my''y'+kyy'=-\gamma(y')^2+F(t)y'.$$

The two terms on the left are derivatives of the two energy terms. Therefore

$$\boxed{\mathcal E'=-\gamma(y')^2+F(t)y'}.$$

This gives a useful check on any solution:

• With no damping or forcing, $\mathcal E'=0$. Kinetic and potential energy exchange, but their sum stays constant.
• With damping and no forcing, $\mathcal E'=-\gamma(y')^2\le0$. Energy is nonincreasing; its derivative is zero at an instant when the velocity is zero.
• The applied force contributes power $F(t)y'$. It can add or remove energy, depending on whether force and velocity have the same sign.

For the preceding example,

$$\mathcal E(0)=\frac12(0.5)(-0.4)^2+\frac12(8)(0.1)^2 =0.08\,\mathrm J.$$

Undamped motion exchanges kinetic and potential energy while the total stays constant.

$y(t)$

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The three damping regimes

For free motion with $\gamma>0$, the characteristic equation and discriminant are

$$mr^2+\gamma r+k=0, \qquad \Delta=\gamma^2-4mk.$$

The roots are

$$r=\frac{-\gamma\pm\sqrt{\Delta}}{2m}.$$

Their form tells us whether the mass swings back and forth.

Overdamped: $\Delta>0$. There are two distinct real roots $r_1,r_2$, and

$$y=C_1e^{r_1t}+C_2e^{r_2t}.$$

Both roots are negative because $\sqrt{\gamma^2-4mk}<\gamma$. The motion decays without a sinusoidal oscillation.

Underdamped: $\Delta<0$. Write the roots as $\lambda\pm i\mu$, where

$$\lambda=-\frac{\gamma}{2m}, \qquad \mu=\frac{\sqrt{4mk-\gamma^2}}{2m} =\sqrt{\frac{k}{m}-\frac{\gamma^2}{4m^2}}.$$

Then

$$y=e^{-\gamma t/(2m)}\bigl(C_1\cos\mu t+C_2\sin\mu t\bigr).$$

The sinusoidal factor produces oscillations; the exponential shrinks their amplitude. The angular frequency $\mu$ is smaller than $\omega_0$. The whole motion is not periodic, because its amplitude changes.

Critically damped: $\Delta=0$. The repeated root gives

$$y=(C_1+C_2t)e^{-\gamma t/(2m)}.$$

The extra factor $t$ does not prevent decay: the negative exponential wins as $t\to\infty$.

All three positively damped free motions tend to equilibrium for every initial displacement and velocity. The threshold separating oscillatory from nonoscillatory motion is

$$\boxed{\gamma_c=2\sqrt{mk}}.$$

For the same mass, stiffness, and release from rest, critical damping gives a faster nonoscillatory return than overdamping. This is useful in shock absorbers, door closers, and measuring needles. It is not a claim that every choice of initial velocity gives monotone motion: even real-root solutions can cross equilibrium.

The same release from rest with damping 0.4, 2, and 5: oscillatory decay, critical return, and slower overdamped return.

$\gamma=0.4$$\gamma=2$$\gamma=5$

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Solving the calibrated spring example

Our earlier equation was $y''+8y'+200y=0$. Its discriminant is $8^2-4(200)=-736$, so its roots are

$$r=-4\pm i\sqrt{184}.$$

With $\mu=\sqrt{184}$, write $y=e^{-4t}(C_1\cos\mu t+C_2\sin\mu t)$. At $t=0$,

$$C_1=0.05,\qquad -4C_1+\mu C_2=0.$$

Thus

$$\boxed{y(t)=e^{-4t}\left(0.05\cos(\sqrt{184}\,t) +\frac{0.2}{\sqrt{184}}\sin(\sqrt{184}\,t)\right)}.$$

The sine coefficient is approximately $0.0147$, but the exact fraction satisfies the initial velocity exactly. The amplitude envelope shrinks by a factor $e^{-4}$ each second. This does not mean $y(1)=e^{-4}y(0)$: the oscillatory factor also changes.

Small damping still matters over long times

Consider

$$y''+0.125y'+y=0, \qquad y(0)=2,\qquad y'(0)=0.$$

The roots are $-1/16\pm i\sqrt{255}/16$. Set $\mu=\sqrt{255}/16$. The initial conditions give $C_1=2$ and $-C_1/16+\mu C_2=0$, so

$$\boxed{y(t)=e^{-t/16}\left(2\cos\frac{\sqrt{255}t}{16} +\frac{2}{\sqrt{255}}\sin\frac{\sqrt{255}t}{16}\right)}.$$

Without damping, the same initial conditions give $2\cos t$. The damped frequency is about $0.998$, nearly $1$, so the curves initially rise and fall at almost the same rate. Their long-term behavior is completely different: only the damped solution tends to zero.

The sine–cosine combination has amplitude $32/\sqrt{255}$, so an exact bound is

$$|y(t)|\le\frac{32}{\sqrt{255}}e^{-t/16}.$$

For instance, this envelope guarantees $|y(t)|<0.1$ once

$$t>16\ln\left(\frac{320}{\sqrt{255}}\right)\approx47.96.$$

The actual solution enters that band permanently a little earlier, at about $47.51$. The envelope gives a safe bound without locating the last crossing.

Small damping compared with no damping, exact exponential bounds, and the inward position–velocity spiral.

$y(t)$$2\cos t$$-\frac{32}{\sqrt{255}}e^{-t/16}$$\frac{32}{\sqrt{255}}e^{-t/16}$

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Plotting the pair $(y(t),y'(t))$ instead of displacement against time gives the inward spiral shown below the time plot. Each point records both position and velocity at the same instant; both tend to zero. That is all we need from this picture here.

Within the underdamped regime, increasing $\gamma$ at fixed $m,k$ makes the envelope decay faster but lowers the oscillation frequency. Faster decay and faster oscillation are different things.

Periodic forcing and steady-state motion

Now apply a periodic force

$$F(t)=F_0\cos(\omega t),$$

where $F_0>0$ is the force amplitude and $\omega>0$ is the forcing angular frequency. This frequency comes from the input; it need not equal the system's natural frequency.

For

$$my''+\gamma y'+ky=F_0\cos(\omega t),$$

try a particular solution $y_p=A\cos(\omega t)+B\sin(\omega t)$. If $\gamma>0$, the homogeneous modes decay, so a pure sine or cosine cannot overlap them. The trial works without an extra factor of $t$.

Example: separating the transient from the forced response

Solve

$$y''+y'+1.25y=3\cos t, \qquad y(0)=2,\qquad y'(0)=3.$$

Substitute $y_p=A\cos t+B\sin t$ and match coefficients:

$$\begin{cases} \frac14A+B=3,\\ -A+\frac14B=0. \end{cases}$$

The second equation gives $B=4A$, so $A=12/17$ and $B=48/17$. Therefore

$$y_p=\frac{12}{17}\cos t+\frac{48}{17}\sin t.$$

Its amplitude is $\sqrt{(12/17)^2+(48/17)^2}=12/\sqrt{17}$. The characteristic roots of the homogeneous equation are $-1/2\pm i$, giving

$$y_c=e^{-t/2}(C_1\cos t+C_2\sin t).$$

Apply the initial conditions to the full sum $y=y_c+y_p$:

$$C_1+\frac{12}{17}=2, \qquad -\frac12C_1+C_2+\frac{48}{17}=3.$$

This gives $C_1=22/17$ and $C_2=14/17$, hence

$$\boxed{ \begin{aligned} y(t)&=\underbrace{e^{-t/2}\left(\frac{22}{17}\cos t+\frac{14}{17}\sin t\right)}_{\text{transient}}\\ &\quad+\underbrace{\frac{12}{17}\cos t+\frac{48}{17}\sin t}_{\text{steady-state response}}. \end{aligned} }$$

The transient adjusts the solution to the starting conditions, then decays. The steady-state response keeps oscillating at the forcing frequency. “Steady” does not mean constant, and the sinusoid still crosses zero. The convergence statement is $y(t)-y_p(t)\to0$, not that $y(t)$ tends to a fixed number.

Total motion approaches the periodic steady state as the transient disappears.

$y(t)$$y_p(t)$$y_c(t)$

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Large response with light damping

Compare the previous model with

$$y''+0.125y'+y=3\cos t, \qquad y(0)=2,\qquad y'(0)=0.$$

The forcing is unchanged, but damping is smaller and the stiffness coefficient is now $1$ rather than $1.25$. The same trial gives

$$0.125B=3,\qquad -0.125A=0,$$

so $y_p=24\sin t$. Its displacement amplitude is $24$: eight times the static displacement $F_0/k=3$. Comparing with a static displacement keeps the units consistent; force and displacement themselves are not the same kind of quantity.

The transient still decays as $e^{-t/16}$, so the response remains bounded and approaches $24\sin t$. For the plotted IVP, the full solution is

$$y(t)=e^{-t/16}\left(2\cos\mu t-\frac{382}{\sqrt{255}}\sin\mu t\right)+24\sin t, \qquad \mu=\frac{\sqrt{255}}{16}.$$

The coefficient follows from $y'(0)=-2/16+\mu C_2+24=0$. Here the forcing frequency equals the undamped natural frequency. The large response is often described as near resonance, but positive damping prevents the unbounded growth of the next example. The exact damped amplitude-peak frequency need not equal $\omega_0$.

Light damping allows a large bounded response. Force divided by stiffness is plotted for a dimensionally consistent comparison.

$y(t)$$y_p(t)$$F(t)/k=3\cos t$

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Undamped resonance

Set $\gamma=0$ and force the system at its natural frequency, $\omega=\omega_0=\sqrt{k/m}$. Then

$$my''+ky=F_0\cos(\omega_0t).$$

Both $\cos(\omega_0t)$ and $\sin(\omega_0t)$ are homogeneous solutions. The ordinary sinusoidal trial produces zero, not the forcing. This does not make undetermined coefficients unusable: multiply the trial by $t$:

$$y_p=At\cos(\omega_0t)+Bt\sin(\omega_0t).$$

Using $k=m\omega_0^2$, substitution simplifies to

$$my_p''+ky_p =2m\omega_0\bigl(B\cos(\omega_0t)-A\sin(\omega_0t)\bigr).$$

Matching the input gives $A=0$ and $B=F_0/(2m\omega_0)$. Thus

$$\boxed{y(t)=C_1\cos(\omega_0t)+C_2\sin(\omega_0t) +\frac{F_0}{2m\omega_0}t\sin(\omega_0t)}.$$

The homogeneous terms stay bounded, but the forced term has an envelope growing linearly in time. No initial conditions can cancel that long-term growth.

For example,

$$y''+y=\frac12\cos t, \qquad y(0)=0,\qquad y'(0)=0$$

has solution $y=t\sin t/4$, bounded at each time by the lines $y=\pm t/4$.

Undamped resonant motion grows between the two straight envelope lines.

$y(t)$$-t/4$$t/4$

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The displacement is not monotonically increasing: it keeps crossing equilibrium while its excursions grow. Resonance builds up energy over time; the instantaneous input power need not be positive at every moment.

This explains why a periodic input near a natural frequency can matter in a vibrating glass, marching on a bridge, or earthquake-resistant design. Resonance is also useful in musical instruments and radio tuning. Real systems have damping and limits to the linear model, so an ideal unbounded solution is not a prediction of literally infinite physical motion.

The Tacoma Narrows Bridge collapse is a different warning: its wind-driven oscillation involved aeroelastic flutter, not simply a prescribed periodic force matching a linear oscillator's natural frequency.

Series RLC circuits

A series circuit contains a resistor, an inductor, and a capacitor in one loop. Let $Q(t)$ be the signed charge on the chosen capacitor plate and choose the current direction so that

$$I(t)=Q'(t).$$

The symbols and units are:

Here $L$ means inductance, not the spring's static extension. Likewise, capacitance $C$ is different from the unit symbol $\mathrm C$ for coulombs.

From voltage drops to an ODE

The three components have voltage drops

$$V_R=RI,\qquad V_C=\frac QC,\qquad V_L=LI'.$$

The resistor dissipates energy as heat. The capacitor stores energy in an electric field between its plates. The inductor stores energy in a magnetic field and opposes changes in current.

Kirchhoff's voltage law says the drops around the loop add to the applied voltage:

$$LI'+RI+\frac QC=E(t).$$

Use $I=Q'$ to obtain the charge equation:

$$\boxed{LQ''+RQ'+\frac1C Q=E(t)}, \qquad Q(t_0)=Q_0,\qquad Q'(t_0)=I_0.$$

Every term has units of volts. For continuous $E$ and constant $L,C>0$, $R\ge0$, initial charge and current determine the solution uniquely.

If we prefer an equation for current and $E$ is continuously differentiable, differentiate once:

$$\boxed{LI''+RI'+\frac1C I=E'(t)}.$$

The initial derivative of current is not an extra independent choice. The original charge equation requires

$$\boxed{I'(t_0)=\frac{E(t_0)-RI_0-Q_0/C}{L}}.$$

Keeping this condition prevents the differentiated equation from introducing a solution inconsistent with the original circuit.

For a constant battery $E(t)=E_0$ and $R>0$, the current eventually tends to zero and $Q(t)\to CE_0$. The charge settles at the value whose capacitor voltage balances the battery. Without resistance, the transient need not decay.

The mechanical–electrical correspondence

Compare the charge equation with $my''+\gamma y'+ky=F(t)$:

These are corresponding mathematical roles, not equal physical quantities. The inductor resists a change in current, much as mass resists a change in velocity; the capacitor supplies the restoring term.

Natural frequency, damping, and energy

For the unforced circuit, the characteristic equation is

$$Lr^2+Rr+\frac1C=0.$$

The natural angular frequency, discriminant, and critical resistance follow directly:

$$\boxed{\omega_0=\frac1{\sqrt{LC}}}, \qquad \Delta=R^2-\frac{4L}{C}, \qquad \boxed{R_c=2\sqrt{\frac LC}}.$$

• $R=0$: undamped natural oscillation.
• $0<R<R_c$: underdamped oscillation, often called ringing.
• $R=R_c$: critical damping, with a repeated real root.
• $R>R_c$: overdamping, with two negative real roots.

These classifications describe the natural response or transient, not every possible externally driven motion. For underdamping, the roots are

$$r=-\frac{R}{2L}\pm i\mu, \qquad \mu=\sqrt{\frac1{LC}-\frac{R^2}{4L^2}}.$$

The stored energy is

$$\mathcal U=\frac12LI^2+\frac{Q^2}{2C}.$$

Multiply the voltage equation by $I=Q'$ to find its derivative:

$$\boxed{\mathcal U'=-RI^2+E(t)I}.$$

The source supplies power $EI$, while $RI^2$ is the nonnegative power converted to heat. No applied voltage does not mean no stored energy.

Why an ideal LC circuit oscillates

With no resistor and no applied voltage, take $Q(0)=Q_0$ and $I(0)=0$. The solution is

$$Q(t)=Q_0\cos(\omega_0t), \qquad I(t)=-\omega_0Q_0\sin(\omega_0t).$$

Let $T=2\pi/\omega_0$. The energy moves between two places during each cycle:

When the capacitor's charge reaches zero, current is still flowing. The inductor's stored energy keeps the current going, charging the capacitor with the opposite polarity. With no resistance, none of the stored energy is converted to heat, so the cycle repeats.

Example: charge in microcoulombs, current in milliamperes

Let

$$L=1\,\mathrm H,\qquad C=0.25\times10^{-6}\,\mathrm F, \qquad Q(0)=10^{-6}\,\mathrm C,\qquad I(0)=0.$$

Then

$$Q''+4\times10^6Q=0, \qquad \omega_0=2000\,\mathrm{rad/s}.$$

With $t$ in seconds,

$$\boxed{Q(t)=10^{-6}\cos(2000t)\quad\mathrm C}, \qquad \boxed{I(t)=-2\times10^{-3}\sin(2000t)\quad\mathrm A}.$$

The period is $\pi/1000\,\mathrm s\approx3.14\,\mathrm{ms}$. The current amplitude is $2\,\mathrm{mA}$, and the total stored energy is

$$\mathcal U(0)=\frac{(10^{-6})^2}{2(0.25\times10^{-6})} =2\times10^{-6}\,\mathrm J.$$

Charge and current exchange electric and magnetic energy with no heat loss.

$Q/(1\,\mu\mathrm C)$$I/(2\,\mathrm{mA})$

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The plot uses milliseconds horizontally, charge in microcoulombs, and current divided by $2\,\mathrm{mA}$ vertically. That normalization makes the quarter-cycle shift visible without confusing a charge with a current.

Adding resistance

Example: an overdamped discharge

Let

$$L=0.2\,\mathrm H,\quad R=300\,\Omega,\quad C=10^{-5}\,\mathrm F, \quad E(t)=0,$$

with $Q(0)=10^{-6}\,\mathrm C$ and $I(0)=0$. The charge equation is

$$0.2Q''+300Q'+10^5Q=0,$$

or, after dividing by $0.2$,

$$Q''+1500Q'+5\times10^5Q=0.$$

The roots are $-500$ and $-1000$. Also, $R_c=2\sqrt{0.2/10^{-5}}\approx282.84\,\Omega$, so $R>R_c$ confirms overdamping.

Write $Q=C_1e^{-500t}+C_2e^{-1000t}$. The initial conditions give

$$C_1+C_2=10^{-6},\qquad -500C_1-1000C_2=0,$$

so $C_1=2\times10^{-6}$ and $C_2=-10^{-6}$. Thus, with $t$ in seconds,

$$\boxed{Q(t)=10^{-6}(2e^{-500t}-e^{-1000t})\quad\mathrm C},$$ $$\boxed{I(t)=10^{-3}(e^{-1000t}-e^{-500t})\quad\mathrm A}.$$

For these initial conditions, the charge decreases to zero without reversing sign. The negative current means the chosen plate is discharging. Its maximum magnitude is $0.25\,\mathrm{mA}$, reached at $t=(\ln2)/500\,\mathrm s\approx1.386\,\mathrm{ms}$; differentiating $I$ locates this point.

Choosing critical resistance

If $L=0.2\,\mathrm H$ and $C=0.8\times10^{-6}\,\mathrm F$, choose

$$R=R_c=2\sqrt{\frac{0.2}{0.8\times10^{-6}}} =1000\,\Omega.$$

This puts the circuit at the boundary between oscillatory and nonoscillatory natural responses. As with the spring, claims about a fastest return should keep the initial conditions and other parameters fixed.

Example: ringing and heat

Return to $L=1\,\mathrm H$ and $C=0.25\,\mu\mathrm F$, but now use $R=400\,\Omega$. Since $R_c=4000\,\Omega$, the circuit is underdamped. With the same initial charge $10^{-6}\,\mathrm C$ and zero initial current,

$$Q''+400Q'+4\times10^6Q=0.$$

The roots have a fixed negative real part:

$$r=-200\pm i\mu,\qquad \mu=\sqrt{3960000}.$$

Applying the initial conditions to $Q=e^{-200t}(C_1\cos\mu t+C_2\sin\mu t)$ gives $C_1=10^{-6}$ and $\mu C_2=200C_1$. Therefore

$$\boxed{Q(t)=10^{-6}e^{-200t}\left(\cos\mu t+\frac{200}{\mu}\sin\mu t\right)}.$$

The amplitude envelope falls by a factor $e^{-1}$ every $1/200\,\mathrm s=5\,\mathrm{ms}$. The charge and current oscillate while tending to zero. The energy is not simply disappearing: it becomes heat in the resistor.

With no external source, let the accumulated heat since time zero be

$$H(t)=\int_0^t RI(s)^2\,ds.$$

The energy balance says

$$\mathcal U(t)+H(t)=\mathcal U(0).$$

Charge and current decay as stored energy becomes heat in the resistor.

$Q/(1\,\mu\mathrm C)$$-\frac{2000}{\sqrt{3960000}}e^{-200t}$$\frac{2000}{\sqrt{3960000}}e^{-200t}$$I/(2\,\mathrm{mA})$

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The stored electric and magnetic energies shrink as heat accumulates. The amplitude decay rate is not the instantaneous energy decay rate: the latter depends on the current through $\mathcal U'=-RI^2$.

The same distinction connects both systems: free motion describes how stored energy evolves, while forced motion also depends on energy exchanged with an outside source. Once the equation and initial conditions are right, the solution methods are the same.