The harmonic motion we discussed here is not physically realistic because it does not take damping due to friction into account. In this note, we discuss *damped harmonic motion* which is physically more realistic. If $x$ is the displacement from the equilibrium position, then the restoring force exerted by a spring is $-kx$ and the retarding force is $-cv$ (friction is proportional to the velocity $v$). Consequently we have the equation $$F=-kx-cv=-kx-c\dot{x}$$ or equivalently the second-order linear equation \begin{equation}\label{eq:damped}m\ddot{x}+c\dot{x}+kx=0\end{equation} To solve the equation \eqref{eq:damped} we still attempt to use the trial solution $x=e^{qt}$. As a result we obtain \begin{equation}\label{eq:chareq}mq^2+cq+k=0\end{equation} \eqref{eq:chareq} is called the *auxiliary equation* or the *characteristic equation*. Its solution is given by $$q=\frac{-c\pm\sqrt{c^2-4mk}}{2m}$$ There are three physically distinct cases:

- $c^2>4mk$:
*overdamping* - $c^2=4mk$:
*critically damping* - $c^2<4mk$:
*underdamping*

*Case 1*. Let $-\gamma_1<0$ and $-\gamma_2<0$ be two real values of $q$. Then the general solution (we will discuss, using linear algebra, why this is indeed the general solution later) is $$x=A_1e^{-\gamma_1 t}+A_2e^{-\gamma_2 t}$$ The motion is nonoscillatory and the displacement $x$ decays to 0 in an exponential matter.

*Example*. The second-order linear differential equation $\ddot{x}+3\dot{x}+2x=0$ has the general solution $$x(t)=A_1e^{-2t}+A_2e^{-t}$$

*Case 2*. $q$ has one real value $q=-\gamma$ where $\gamma=\frac{c}{2m}$. The equation \eqref{eq:damped} can be written as \begin{equation}\label{eq:criticdamped}\left(\frac{d}{dt}+\gamma\right)^2x=\left(\frac{d}{dt}+\gamma\right)\left(\frac{d}{dt}+\gamma\right)x=0\end{equation} Let $u=\frac{d}{dt}+\gamma$. Then \eqref{eq:criticdamped} reduces to a first-order differential equation $$\left(\frac{d}{dt}+\gamma\right)u=\frac{du}{dt}+\gamma u=0$$ which is separable. It’s solution is $u=A_1e^{-\gamma t}$. Now we have a first-order linear differential equation $$\frac{dx}{dt}+\gamma x=A_1e^{-\gamma t}$$ whose solution is given by $$x(t)=e^{-\gamma t}(A_1 t+A_2)$$

*Example*. The second-order linear differential equation $$\ddot{x}+2\dot{x}+x=0$$ has the general solution $$x(t)=e^{-t}(A_1t+A_2)$$

This also represents a nonoscillatory motion and the displacement $x$ decays to zero asymptotically. Critical damping produces an optimal return to the equilibrium position, so it is used, for example, for galvanometer suspensions.

*Case 3*. Suppose that $c$ is small enough so that $c^2-4mk<0$. In this case, $q$ are two complex numbers $-\gamma\pm i\omega_1$ where $\gamma=\frac{c}{2m}$, $\omega_1=\sqrt{\frac{k}{m}-\frac{c^2}{4m^2}}=\sqrt{\omega_0^2-\gamma^2}$. So $e^{(-\gamma+i\omega_1)t}$, $e^{(-\gamma-i\omega_1)t}$ are solutions of \eqref{eq:damped}. Due to the linearity of \eqref{eq:damped}, the real part $e^{-\gamma t}\cos\omega_1 t$and the imaginary part $e^{-\gamma t}\sin\omega_1 t$ of $e^{(-\gamma+i\omega_1)t}$ are also solutions of \eqref{eq:damped}. Hence, $$x(t)=ae^{-\gamma t}\cos\omega_1 t+be^{-\gamma t}\sin\omega_1 t$$ is the general solution. This can be written as $$x(t)=Ae^{-\gamma t}\cos(\omega_1 t-\theta_0)$$ where $A=\sqrt{a^2+b^2}$ and $\theta_0=\tan^{-1}\left(\frac{b}{a}\right)$. Note that the angular frequency (or natural frequency) $\omega_1$ is smaller than that of undamped harmonic oscillator $\omega_0$. $$\omega_1=\sqrt{\omega_0^2-\gamma^2}=\omega_0\sqrt{1-\left(\frac{\gamma}{\omega_0}\right)^2}$$ If $\frac{\gamma}{\omega_0}<1$ then $\sqrt{1-\left(\frac{\gamma}{\omega_0}\right)^2}\approx 1-\frac{1}{2}\frac{\gamma^2}{\omega_0^2}$. So $$\omega_1\approx \omega_0-\frac{\gamma^2}{2\omega_0}$$

*Example*. Let $m=1$, $c=4$ and $k=404$. Then the resulting equation of damped harmonic oscillator is $$\ddot{x}+4\dot{x}+404=0$$ Let us solve this equation with $x(0)=1$ and $\dot{x}(0)=0$. The characteristic equation $$q^2+4q+404=0$$ has two complex solutions $q=-2\pm 20i$. Thus $$x(t)=ae^{-2t}\cos(20t)+be^{-2t}\sin(20t)$$ $x(0)=1$ results in $a=1$. To determine $b$ we need $\dot{x}$. $$\dot{x}=-2e^{-2t}\cos(20t)-20e^{-2t}\sin(20t)-2be^{-2t}\sin(20t)+20be^{-2t}\cos(20t)$$ The initial condition $\dot{x}(0)=0$ results in $b=\frac{1}{10}$. Hence $x(t)$ is given by $$x(t)=e^{-2t}\cos(20t)+\frac{1}{10}e^{-2t}\sin(20t)$$ Since $A=\sqrt{a^2+b^2}=\frac{\sqrt{101}}{10}$ and $\theta_0=\tan^{-1}\left(\frac{b}{a}\right)=\tan^{-1}\left(\frac{1}{10}\right)\approx 0.099669$, $x(t)$ also can be written as $$x(t)=\frac{\sqrt{101}}{10}e^{-2t}\cos(20t-0.099669)$$

Differentiate the total energy $$E=\frac{1}{2}m\dot{x}^2+\frac{1}{2}kx^2$$ with respect to $t$. \begin{align*}\frac{dE}{dt}&=m\ddot{x}\dot{x}+k\dot{x}x\\&=(m\ddot{x}+kx)\dot{x}\\&=(-c\dot{x})\dot{x}\\&=-c(\dot{x})^2<0\end{align*} This is the rate at which energy is dissipated into heat by friction.

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