Damped Harmonic Oscillation

Consider the mass–spring system discussed in Section 1.2. Suppose that, as it slides over the horizontal surface, the mass is subject to a frictional damping force that opposes its motion, and is directly proportional to its instantaneous velocity. It follows that the net force acting on the mass when its instantaneous displacement is $x(t)$ takes the form

$$f (x,\dot{x})= - k\,x -m\,\nu\,\dot{x},$$ (2.1)

where $m>0$ is the mass, $k>0$ the spring force constant, and $\nu>0$ a constant (with the dimensions of angular frequency) that parameterizes the strength of the damping. The time evolution equation of the system thus becomes [cf., Equation (1.2)]

$$\ddot{x} + \nu\,\dot{x} + \omega_0^{\,2}\,x = 0,$$ (2.2)

where $\omega_0=\sqrt{k/m}$ is the undamped oscillation frequency [cf., Equation (1.6)]. We shall refer to the preceding equation as the damped harmonic oscillator equation.

It is worth discussing the two forces that appear on the right-hand side of Equation (2.1) in more detail. The first is the restoring force that develops when a mechanical system in a stable equilibrium state is slightly disturbed from that state. We can suppose, without loss of generality, that the equilibrium state corresponds to $x=0$. Suppose that $f(x)$ is the restoring force. It follows that $f(0)=0$ and $df(0)/dx<0$, otherwise $x=0$ is not a stable equilibrium point. Taylor expanding about $x=0$ (Riley 1974), we obtain

$$f(x) \simeq -k\,x+\alpha\,x^{\,2} + {\cal O}\left(x^{\,3}\right),$$ (2.3)

where $k=-df(0)/dx$, and $\alpha=2\,d^{\,2}f(0)/dx^{\,2}$. It follows that, provided the magnitude of the displacement from the equilibrium point, $\vert x\vert$, remains sufficiently small, we can always approximate $f(x)$ by the first non-zero term in its Taylor expansion; that is, $f(x)\simeq -k\,x$. Thus, the first term that appears on the right-hand side of Equation (2.1) is exact (as long as the amplitude of the motion remains sufficiently small). The second term, on the other hand, which represents the damping force, is completely phenomenological. In other words, damping forces that arise in nature are not necessarily directly proportional to the instantaneous velocity of the system. For example, the damping force that arises from air resistance is usually directly proportional to the square of the instantaneous velocity (Batchelor 2000). In fact, the only reason that we have chosen a damping force that is directly proportional to the instantaneous velocity is that it leads to a linear equation of motion.

Let us search for a solution to Equation (2.2) of the form

$$x(t) = a\,{\rm e}^{-\gamma\,t}\,\cos(\omega_1\,t-\phi),$$ (2.4)

where $a>0$, $\gamma>0$, $\omega_1>0$, and $\phi$ are all constants. By analogy with the discussion in Section 1.2, we can interpret the preceding solution as a periodic oscillation, of fixed angular frequency $\omega_1$, and phase angle $\phi$, whose amplitude decays exponentially in time as $a(t)=a\,\exp(-\gamma\,t)$. It can be demonstrated that

$$\dot{x} = - \gamma\,a\,{\rm e}^{-\gamma\,t}\,\cos(\omega_1\,t-\phi) -\omega_1\, a\,{\rm e}^{-\gamma\,t}\,\sin(\omega_1\,t-\phi),$$ (2.5)

$$\ddot{x} = \left(\gamma^{\,2}-\omega_1^{\,2}\right)a\,{\rm e}^{-\gamma\,t}... ...t-\phi) +2\,\gamma\,\omega_1\, a\,{\rm e}^{-\gamma\,t}\,\sin(\omega_1\,t-\phi).$$ (2.6)

Hence, collecting similar terms, Equation (2.2) becomes

$$=\left[ \left(\gamma^{\,2}-\omega_1^{\,2}\right) -\nu\,\gamma + \omega_0^{\,2}\right]a\,{\rm e}^{-\gamma\,t}\,\cos(\omega_1\,t-\phi)$$ 0

$$~~+\left(2\,\gamma\,\omega_1-\nu\,\omega_1\right) a\,{\rm e}^{-\gamma\,t}\,\sin(\omega_1\,t-\phi).$$ (2.7)

The only way that the preceding equation can be satisfied at all times is if the (constant) coefficients of $\exp(-\gamma\,t)\,\cos(\omega_1\,t-\phi)$ and $\exp(-\gamma\,t)\,\sin(\omega_1\,t-\phi)$ separately equate to zero, so that

$$\left(\gamma^{\,2}-\omega_1^{\,2}\right) -\nu\,\gamma+\omega_0^{\,2} =0,$$ (2.8)

$$2\,\gamma\,\omega_1-\nu\,\omega_1 =0.$$ (2.9)

These equations can be solved to give

$$\gamma= \nu/2,$$ (2.10)

and

$$\omega_1 =\left(\omega_0^{\,2}-\nu^{\,2}/4\right)^{1/2}.$$ (2.11)

Thus, the solution to the damped harmonic oscillator equation is written

$$x(t) = a\,{\rm e}^{-\nu\,t/2}\,\cos\left(\omega_1 \,t-\phi\right),$$ (2.12)

assuming that $\nu< 2\,\omega_0$ (because $\omega_1^{\,2}=\omega_0^{\,2}-\nu^{\,2}/4$ cannot be negative). We conclude that the effect of a relatively small amount of damping, parameterized by the damping constant, $\nu$, on a system that exhibits simple harmonic oscillation about a stable equilibrium state is to reduce the angular frequency of the oscillation from its undamped value $\omega_0$ to $\omega_1=(\omega_0^{\,2}-\nu^{\,2}/4)^{1/2}$, and to cause the amplitude of the oscillation to decay exponentially in time at the rate $\nu/2$. This modified type of oscillation, which we shall refer to as damped harmonic oscillation, is illustrated in Figure 2.1. Incidentally, if the damping is sufficiently large that $\nu\geq 2\,\omega_0$ (which we shall assume not to be the case) then the system does not oscillate at all, and any motion simply decays away exponentially in time. (See Exercise 7.)

Figure 2.1: Damped harmonic oscillation. Here, $T_0=2\pi /\omega _0$, $\nu\,T_0=0.5$, and $\phi = 0$. The solid line shows $x(t)/a$, whereas the dashed lines show $\pm a(t)/a$.

Although the angular frequency, $\omega_1$, and decay rate, $\nu/2$, of the damped harmonic oscillation specified in Equation (2.12) are determined by the constants appearing in the damped harmonic oscillator equation, (2.2), the initial amplitude, $a$, and the phase angle, $\phi$, of the oscillation are determined by the initial conditions. In fact, if $x(0)=x_0$ and $\dot{x}(0)=v_0$ then it follows from Equation (2.12) that

$$x_0 =a\,\cos\phi,$$ (2.13)

$$v_0 = - \frac{\nu}{2}\,a\,\cos\phi + \omega_1\,a\,\sin\phi = -\frac{\nu\,x_0}{2}+ \omega_1\,a\,\sin\phi,$$ (2.14)

which implies that

$$a\,\cos\phi = x_0,$$ (2.15)

$$a\,\sin\phi = \frac{v_0 + \nu\,x_0/2}{\omega_1},$$ (2.16)

giving

$$a =\left[x_0^{\,2} + \left(\frac{v_0+\nu\,x_0/2}{\omega_1}\right)^2\right]^{1/2},$$ (2.17)

$$\phi = \tan^{-1}\left(\frac{v_0+\nu\,x_0/2}{\omega_1\,x_0}\right).$$ (2.18)

The damped harmonic oscillator equation is a linear differential equation. In other words, if $x(t)$ is a solution then so is $c\,x(t)$, where $c$ is an arbitrary constant. It follows that the solutions of this equation are superposable, so that if $x_1(t)$ and $x_2(t)$ are two solutions corresponding to different initial conditions then $\alpha\,x_1(t)+\beta\,x_2(t)$ is a third solution, where $\alpha$ and $\beta$ are arbitrary constants.

Multiplying the damped harmonic oscillator equation, (2.2), by $\dot{x}$, we obtain

$$\dot{x}\,\ddot{x} + \nu\,\dot{x}^{\,2}+ \omega_0^{\,2}\,\dot{x}\,x=0,$$ (2.19)

which can be rearranged to give

$$\frac{dE}{dt} = - m\,\nu\,\dot{x}^{\,2},$$ (2.20)

where

$$E = \frac{1}{2}\,m\,\dot{x}^{\,2}+\frac{1}{2}\,k\,x^{\,2}$$ (2.21)

is the total energy of the system; that is, the sum of the kinetic and potential energies. Because the right-hand side of (2.20) cannot be positive, and is only zero when the system is stationary, the total energy is not a conserved quantity, but instead decays monotonically in time due to the action of the damping. The net rate at which the force (2.1) does work on the mass is

$$P = f\,\dot{x} = -k\,\dot{x}\,x-m\,\nu\,\dot{x}^{\,2}.$$ (2.22)

The spring force (i.e., the first term on the right-hand side) does negative work on the mass (i.e., it reduces the system kinetic energy) when $\dot{x}$ and $x$ are of the same sign, and does positive work when they are of the opposite sign. It can easily be demonstrated that, on average, the spring force does no net work on the mass during an oscillation cycle. The damping force, on the other hand, (i.e., the second term on the right-hand side) always does negative work on the mass, and, therefore, always acts to reduce the system kinetic energy.