This chapter has three objectives:

1. understand the damped oscillator;

2. learn to solve linear, homogeneous differential equations;

3. use trigonometric identities to rearrange formulas.

4.1 Recall: the RC circuit

For the discharging capacitor, (2.3) reads (with a dot for \(d/dt\)) \[\dot Q(t) = -\frac{1}{RC}\,Q(t).\] Guess an exponential, \(Q(t) = Q_0\,e^{\lambda t}\). Substituting gives \[\keyeq{\displaystyle \lambda = -\frac{1}{RC}} \qquad\therefore\qquad \keyeq{\displaystyle Q(t) = Q_0\,e^{-t/RC}}\]

4.2 Simple harmonic oscillator

Equation of motion

Newton’s second law, \(F = ma\) with \(F = F_s\), gives \(ma = -kx\), that is \[\begin{equation} \tag{4.1} \keyeq{\displaystyle \frac{d^2x}{dt^2} + \frac{k}{m}\,x = 0} \end{equation}\] This is a second-order differential equation.

Solution

Use the same strategy as for the RC circuit. Guess \(x(t) = A\,e^{\lambda t}\), so \[\frac{dx}{dt} = \lambda A\,e^{\lambda t}, \qquad \frac{d^2x}{dt^2} = \lambda^2 A\,e^{\lambda t}.\] Plugging into (4.1) gives \(\lambda^2 x + \dfrac{k}{m}\,x = 0\), so \[\keyeq{\displaystyle \lambda = \pm\sqrt{-\frac{k}{m}}}\] There are two solutions (two derivatives), and \(\sqrt{-1} = i\) is imaginary! Call \(\lambda = i\omega\), with \[\omega = \sqrt{\frac{k}{m}} .\] Then \[\begin{equation} \tag{4.2} \keyeq{x(t) = A\,e^{+i\omega t} + B\,e^{-i\omega t}} \end{equation}\] which we call Form 1. How do we interpret it?

Euler’s formula

Use Euler’s formula \[\keyeq{e^{i\omega t} = \cos\omega t + i\sin\omega t}\] Then \[\begin{align*} x(t) &= A\,(\cos\omega t + i\sin\omega t) + B\,(\cos\omega t - i\sin\omega t) \\ &= \underbrace{(A+B)}_{\equiv\, C}\cos\omega t + \underbrace{i\,(A-B)}_{\equiv\, D}\sin\omega t \end{align*}\] \[\begin{equation} \tag{4.3} \keyeq{x(t) = C\cos\omega t + D\sin\omega t} \end{equation}\] which is Form 2. Here \(\omega\) is the angular frequency of the oscillations, and \(C\), \(D\) are set by the initial conditions. Why? Because \[\keyeq{x(t=0) = C} \qquad v(t) = -C\omega\sin\omega t + D\omega\cos\omega t \;\Longrightarrow\; \keyeq{v(t=0) = D\,\omega}\]

Frequency and period

\[\omega = 2\pi f, \qquad T = \frac{1}{f} = \frac{2\pi}{\omega},\] where \(f\) is the frequency of the oscillations and \(T\) the period.

One more identity

Use \(\cos(a+b) = \cos a\cos b - \sin a\sin b\): \[E\cos(\omega t + \phi) = \underbrace{(E\cos\phi)}_{\equiv\, C}\cos\omega t - \underbrace{(E\sin\phi)}_{\equiv\, D}\sin\omega t .\] Note that \[\keyeq{C^2 + D^2 = E^2\cos^2\phi + E^2\sin^2\phi = E^2} \qquad \keyeq{\displaystyle -\frac{D}{C} = \frac{\sin\phi}{\cos\phi} = \tan\phi}\] Therefore \[\begin{equation} \tag{4.4} \keyeq{x(t) = E\cos(\omega t + \phi)} \end{equation}\] which is Form 3: a shifted oscillating function with amplitude \(E\).

4.3 Damped oscillator

Newton’s second law, \(ma = -kx - bv\), gives \[\begin{equation} \tag{4.5} \keyeq{\displaystyle m\,\frac{d^2x}{dt^2} + b\,\frac{dx}{dt} + k\,x = 0} \end{equation}\]

Solution

Guess \(x(t) = A\,e^{\lambda t}\). Then (4.5) becomes \(m\lambda^2 + b\lambda + k = 0\), so \[\lambda = -\frac{b}{2m} \pm \sqrt{\left(\frac{b}{2m}\right)^2 - \frac{k}{m}} .\] Call \[\tau = \frac{2m}{b}, \qquad \omega_0 = \sqrt{\frac{k}{m}} .\] Then if \(\omega_0^2 > 1/\tau^2\) the square root is imaginary and \[\begin{align*} x(t) &= e^{-t/\tau}\left(A\,e^{i\omega t} + B\,e^{-i\omega t}\right) &&\text{(Form 1)} \\ &= \keyeq{E\,e^{-t/\tau}\cos(\omega t + \phi)} &&\text{(Form 3)} \end{align*}\] where \[\omega = \sqrt{\frac{k}{m} - \left(\frac{b}{2m}\right)^2} .\] This is a damped oscillator.

Problems

Problems are taken from the sources listed under Sources and Attribution. Numbers refer to the end-of-chapter problems; figures have been redrawn.

OpenStax University Physics Vol. 1, Chapter 15: Oscillations

OpenStax Calculus Vol. 3, Chapter 7: Second-Order Differential Equations