Constructing the Spring-Mass System
We will start from scratch where a mass \(m\) is connected to and hung by a spring vertically in mid-air, as demonstrated in the diagram below. The spring has a spring constant \(k\) and the force exerted by it on the mass is \( -kz \) by Hooke’s Law, where \( z \) is the vertical displacement (taking downward as positive) of the spring (stretched/compressed) from its unperturbed position. Then by Newton’s Second Law, the governing equation is
\begin{equation}
F = -kz + mg = m\frac{d^2z}{dt^2} \tag{1}
\end{equation}
with \( g \) as the gravitational acceleration. We can do a variable transformation \( z = x + mg/k \) so that the expression of the net force acting on the mass
\begin{align}
m\frac{d^2}{dt^2}(x + \frac{mg}{k}) &= -k(x + \frac{mg}{k}) + mg \\
m\frac{d^2x}{dt^2} &= -kx -mg + mg = -kx \tag{2}
\end{align}
is simplified to being linearly proportional to the equilibrium position \( x \), i.e. \( F=0 \) when \( x=0 \) (\( z = mg/k \)).
The spring-mass system (2) \( d^2x/dt^2 = -(k/m)x \) has a simple general solution of
\begin{align}
x &= c_1 \cos ((\sqrt{k/m}) t) + c_2 \sin ((\sqrt{k/m}) t) \\
&= c_1 \cos (\omega_0 t) + c_2 \sin (\omega_0 t) \tag{3}
\end{align}
where \( \omega_0 = \sqrt{k/m} \) is referred to as the intrinsic frequency. This kind of sinusoidal motion is called a simple harmonic oscillator.
Now we will add a damper to the spring-mass system. The damping force is assumed to be linearly proportional to the opposite of velocity \( v = dx/dt \) so that
\begin{align}
m\frac{d^2x}{dt^2} &= -kx -b\frac{dx}{dt} \tag{4}
\end{align}
where \( b \) is the damping coefficient. Rearranging it further:
\begin{align}
m\frac{d^2x}{dt^2} + b\frac{dx}{dt} + kx &= 0 \\
\frac{d^2x}{dt^2} + \frac{b}{m}\frac{dx}{dt} + \frac{k}{m}x &= 0 \\
\frac{d^2x}{dt^2} + \alpha\frac{dx}{dt} + \omega_0^2 x &= 0 \tag{5}
\end{align}
where we set \( \alpha = b/m \). We now have the equation of motion for a damped harmonic oscillator as a general second-order constant-coefficient homogeneous ODE. According to previous tutorials, there will be three types of solutions/responses depending on the magnitude of \( \alpha \), and they are under-damped, over-damped, and critically-damped, explained as follows.
Under-damped
If the damping is weak, we may expect that there would still be oscillations but the amplitude has to decay gradually. This is known as under-damped and happens when the discriminant for the auxiliary equation \( r^2 + \alpha r + \omega_0^2 = 0\) is negative, i.e. \( \alpha^2 -4\omega_0^2 < 0 \) or \( \alpha < 2\omega_0 \) (\(\alpha \) and \( \omega_0 \) are always positive). The two complex roots for the auxiliary equation are
\begin{align}
r_{\pm} &= \frac{-\alpha \pm \sqrt{\alpha^2 -4\omega_0^2}}{2} \\
&= -\frac{\alpha}{2} \pm \left(\sqrt{\omega_0^2 -\frac{\alpha^2}{4}}\right) i \\
&= -\frac{\alpha}{2} \pm \omega_1 i \tag{6}
\end{align}
where we let \( \omega_1 = \sqrt{\omega_0^2 -\alpha^2/4}\). The general solution is hence
\begin{equation}
x = c_1 e^{-(\alpha/2) t} \cos(\omega_1 t) + c_2 e^{-(\alpha/2) t} \sin(\omega_1 t) \tag{7}
\end{equation}
By using the trigonometric identity \( \cos\alpha \cos\beta + \sin\alpha\sin\beta = \cos(\alpha -\beta) \Rightarrow \) \( \cos(\phi)\cos(\omega_1 t) + \sin(\phi)\sin(\omega_1 t) = \cos(\omega_1 t -\phi) \), we can reorganize the solution as
\begin{align}
x &= Ae^{-(\alpha/2) t} \cos\phi\cos(\omega_1 t) + Ae^{-(\alpha/2) t} \sin\phi\sin(\omega_1 t) \\
&= Ae^{-(\alpha/2) t} \cos(\omega_1 t -\phi) \tag{8}
\end{align}
where \( c_1 = A\cos\phi, c_2 = A\sin\phi, A = \sqrt{c_1^2+c_2^2}, \tan\phi = c_2/c_1\).
Compared to the simple harmonic oscillator case, the amplitude decays exponentially with a characteristic rate of \( \alpha/2 \). Moreover, the damping causes the actual frequency \( \smash{\omega_1 = \sqrt{\omega_0^2 -\alpha^2/4}}\) to be lower than the intrinsic frequency \( \omega_0 \).
Over-damped
In contrast, when the discriminant is positive, so that \( \alpha^2 -4\omega_0^2 > 0 \) or \( \alpha > 2\omega_0 \), there will be two real roots for the auxiliary equation:
\begin{align}
r_{\pm} &= \frac{-\alpha \pm \sqrt{\alpha^2 -4\omega_0^2}}{2} \\
&= -\frac{\alpha}{2} \pm \sqrt{\frac{\alpha^2}{4} -\omega_0^2} \tag{9}
\end{align}
Assigning \( \alpha_1/2 = -r_+ = \alpha/2 -\smash{\sqrt{\alpha^2/4 -\omega_0^2}} \), \( \alpha_2/2 = -r_- = \alpha/2 + \smash{\sqrt{\alpha^2/4 -\omega_0^2}} \), the general solution will be
\begin{align}
x = c_1e^{-(\alpha_1/2)t} + c_2e^{-(\alpha_2/2)t} \tag{10}
\end{align}
where \( \alpha_1 < \alpha_2 \) and the first part of the solution will decay more slowly than the second part. This is referred to as over-damped.
Critically-damped
The final case is critically-damped where \( \alpha = 2\omega_0 \) exactly and the auxiliary equation has only one repeated real root: \( r = \alpha/2 \). The general solution will then be
\begin{equation}
x = c_1e^{-(\alpha/2)t} + c_2te^{-(\alpha/2)t} \tag{11}
\end{equation}
Given a fixed intrinsic frequency \( \omega_0 \) and a varying damping force \( \alpha \), the decay of displaced amplitude in the critically-damped scenario will be the fastest. To see this, the under-damped case has the same form of decay rate \( \alpha/2 \) but with a smaller \( \alpha < 2\omega_0\) than the critically-damped case \( \alpha = 2\omega_0 \). Meanwhile, in the over-damped case, we can observe that \( \alpha_1/2 = \alpha/2 -\smash{\sqrt{\alpha^2/4 -\omega_0^2}} \leq \omega_0 \), by considering a right-triangle with sides \( \omega_0 \), \( \smash{\sqrt{\alpha^2/4 -\omega_0^2}} \), and the hypotenuse \( \alpha/2 \), so the corresponding part of the over-damped solution will decay slower than that of the critically-damped one.
Exercise
Consider the RLC circuit system:
\begin{equation}
L\frac{d^2q}{dt} + R\frac{dq}{dt} + \frac{1}{C}q = 0 \tag{12}
\end{equation}
where \( q \) is the amount of charge, \( L \) is the inductance, \( R \) is the resistance, and \( C \) is the capacitance. Derive the intrinsic frequency and determine when the critical damping arises by comparing (12) to the above discussion.
Answer
Dividing by \( L \) to get:
\begin{equation}
\frac{d^2q}{dt} + \frac{R}{L}\frac{dq}{dt} + \frac{1}{LC}q = 0
\end{equation}
A direct comparison reveals that \( \alpha = R/L \), \( \omega_0 = \smash{\sqrt{1/LC}} \), and the critical damping occurs when
\begin{align}
\alpha &= 2\omega_0 \\
\frac{R}{L} &= 2\sqrt{\frac{1}{LC}} \\
\frac{R}{2} &= \sqrt{\frac{L}{C}}
\end{align}
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