= Energy dissipation of a damped pendulum
{title2=$\dot E=-c\dot\theta^2$}
For the normalized <damped pendulum>, the sum of <kinetic energy> and gravitational <potential energy>, divided by $mg\ell$, is
$$
E(\tau)=\frac12\dot\theta^2+1-\cos\theta,\qquad
\dot E=\dot\theta(\ddot\theta+\sin\theta)=-c\dot\theta^2\leq0.
$$
The <chain rule> and the full nonlinear equation give this identity exactly, without a <small-angle approximation>. Here $\dot\theta=d\theta/d\tau$ as in <damped pendulum>. In physical units the <mechanical energy> and the <power> of the resisting <torque> are
$$
\mathcal E(t)=\frac12m\ell^2\left(\frac{d\theta}{dt}\right)^2+mg\ell(1-\cos\theta),\qquad
\frac{d\mathcal E}{dt}=-b\left(\frac{d\theta}{dt}\right)^2.
$$
The pendulum's lost <mechanical energy> is transferred to the damper and surroundings; it does not violate <conservation of energy>. In an ideal viscous <dashpot> this loss is <frictional heating>. The exact cumulative balance is particularly useful for checking a numerical solution:
$$
Q(\tau)=c\int_0^\tau\dot\theta(s)^2\,ds,\qquad E(\tau)+Q(\tau)=E(0).
$$
\Image[/damped-pendulum-nonlinear-energy.png]
{title=Illustrative nonlinear and linear damped pendulum solutions at 10 and 60 degrees, with exact mechanical energy transfer and loss; no measured data}
The figure uses $c=0.20$ and release from rest. Over $0\leq\tau\leq25$, the largest angular discrepancy between the nonlinear solution and the <damped harmonic oscillator> solution is approximately $0.000702\,\mathrm{rad}$ ($0.0402^\circ$) for a $10^\circ$ release, but $0.14995\,\mathrm{rad}$ ($8.592^\circ$) for a $60^\circ$ release. Their nonlinear initial energies are $1-\cos10^\circ\approx0.015192$ and $1-\cos60^\circ=0.5$, respectively. At $\tau=25$ the remaining fractions are approximately $0.0064080$ and $0.0063757$. These are illustrative calculations from specified initial conditions, not observations. The generator uses a <Runge-Kutta method> and integrates $Q$ alongside the motion; halving its step and independently integrating $c\dot\theta^2$ checks the result.
For $\ell=1\,\mathrm m$ and $g=9.81\,\mathrm{m/s^2}$, the first three nonlinear same-side swing intervals after a $60^\circ$ release are approximately $2.1008$, $2.0378$, and $2.0221\,\mathrm s$. They approach the linear-model <oscillation period> $2.0162\,\mathrm s$ as the <amplitude> shrinks. After a $10^\circ$ release the corresponding intervals are already about $2.0185$, $2.0168$, and $2.0164\,\mathrm s$. This makes the experiment's period drift a quantitative test of the <small-angle approximation>, rather than an automatic reason to change the damping coefficient.
At a turning point, $\dot\theta=0$ and $\dot E=0$ even though the swing is losing <mechanical energy> over the cycle. Its <potential energy> there is $E_n=1-\cos A_n$, so measured turning <angles> give successive energies without estimating <angular velocity>. In the proposed decay experiment, multiply these values by $mg\ell$ to obtain energies in <joules>, then subtract adjacent same-side energies to estimate the loss over each full swing. Only for small <amplitudes> is $E_n\simeq A_n^2/2$. In the linear underdamped model, same-side peak energies consequently satisfy $E_{n+1}/E_n=e^{-\gamma T_d}$, the square of the angular <amplitude> ratio. For the illustrative parameters above this energy ratio is approximately $0.2828$. Between turning points, <kinetic energy> and <potential energy> continue to exchange while their sum decreases; the energy slope vanishes briefly at each extremum and is negative whenever the bob moves.
It is therefore inaccurate to replace the exact identity by $\dot E=-cE$. Even for the linear model, $E_{\mathrm{lin}}=(\dot\theta^2+\theta^2)/2$ has $\dot E_{\mathrm{lin}}=-c\dot\theta^2$. For weak damping and small <angles>, cycle averaging gives $\langle\dot\theta^2\rangle\simeq\langle E_{\mathrm{lin}}\rangle$, so the averaged energy approximately follows $e^{-c\tau}$ while angular <amplitude> follows $e^{-c\tau/2}$. This <exponential decay> is an approximation to the energy trace; the same-side peak ratio is exact within the linear model. The figure distinguishes the actual energy curves from this averaging approximation.
The energy identity also gives a global conclusion without solving the nonlinear equation. If $E$ were constant on a positive-length interval, then $\dot\theta$ would vanish throughout it; the equation would force $\sin\theta=0$, making the entire trajectory an <equilibrium>. Every nonstationary trajectory therefore loses <mechanical energy> over every positive-length interval, even though its instantaneous <derivative> can vanish at a turning point. A nonconstant periodic motion would have to return to its initial <mechanical energy>, so it is impossible for this undriven <damped pendulum>. Oscillations near a <stable focus> are successive, shrinking swings, not a repeating cycle. Moreover, if $E(0)<2$, the <angle> cannot reach an upright <equilibrium>, whose <potential energy> is $2$; such motion stays within its original potential well. This is the usefulness of the energy description: one scalar balance controls both the observed decay and which nonlinear motions remain accessible.
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