Forced Spring-Mass System
In the previous tutorial, we discussed the damped spring-mass system as a type of harmonic oscillator, without any external force exerted on it. In general, the representing ODE is
\begin{equation}
\frac{d^2x}{dt^2} + \alpha\frac{dx}{dt} + \omega_0^2 x = G(t) \tag{1}
\end{equation}
where the source term \( G(t) = 0 \) was zero before. Now, we will consider the case where a new forcing \( G(t) \neq 0 \) is applied to the system. It is further assumed that the forcing is independent of the state \(x\) while sinusoidal in time \(t\), particularly consists of a single frequency (monochromatic) \( G(t) = (\hat{F}/m)\cos(\omega t) \) for an easier treatment. The ODE then becomes
\begin{equation}
\frac{d^2x}{dt^2} + \alpha\frac{dx}{dt} + \omega_0^2 x = \frac{\hat{F}}{m}\cos(\omega t) \tag{2}
\end{equation}
The techniques introduced in the last few tutorials can be used to derive the particular solution corresponding to this source term. Specifically, the method of Undetermined Coefficients will be more convenient: let’s guess a particular solution in the form of \( x_p(t) = C \cos(\omega t) + S \sin(\omega t) \), then
\begin{align}
x_p’ &= -C\omega \sin(\omega t) + S\omega \cos(\omega t) \tag{3} \\
x_p^{\prime\prime} &= -C\omega^2 \cos(\omega t) -S\omega^2 \sin(\omega t) \tag{4}
\end{align}
Plugging all of these into the ODE (2) results in
\begin{align}
\begin{aligned}
&(-C\omega^2 \cos(\omega t) -S\omega^2 \sin(\omega t)) \\
& +\alpha (-C\omega \sin(\omega t) + S\omega \cos(\omega t)) \\
& +\omega_0^2 (C \cos(\omega t) + S \sin(\omega t))
\end{aligned}
= \frac{\hat{F}}{m} \cos (\omega t) \tag{5}
\end{align}
Grouping the terms and comparing the coefficients for the sine/cosine, we have
\begin{align}
\left\{\begin{aligned}
(\omega_0^2-\omega^2) C + \alpha \omega S &= \frac{\hat{F}}{m} \\
-\alpha \omega C + (\omega_0^2-\omega^2) S &= 0
\end{aligned}\right. \tag{6}
\end{align}
This linear system can be solved as
\begin{align}
\left\{\begin{aligned}
C &= \frac{\hat{F}}{m}\frac{\omega_0^2-\omega^2}{(\omega_0^2-\omega^2)^2 + \alpha^2\omega^2} \\
S &= \frac{\hat{F}}{m}\frac{a\omega}{(\omega_0^2-\omega^2)^2 + \alpha^2\omega^2}
\end{aligned}\right. \tag{7}
\end{align}
Therefore,
\begin{align}
x_p(t) &= C \cos(\omega t) + S \sin(\omega t) \\
&= \frac{\hat{F}}{m} \left(\frac{\omega_0^2-\omega^2}{(\omega_0^2-\omega^2)^2 + \alpha^2\omega^2}\cos(\omega t) + \frac{a\omega}{(\omega_0^2-\omega^2)^2 + \alpha^2\omega^2} \sin(\omega t)\right) \\
&= \frac{\hat{F}/m}{\sqrt{(\omega_0^2-\omega^2)^2 + \alpha^2\omega^2}} \\
&\quad \left(\frac{\omega_0^2-\omega^2}{\sqrt{(\omega_0^2-\omega^2)^2 + \alpha^2\omega^2}}\cos(\omega t) + \frac{a\omega}{\sqrt{(\omega_0^2-\omega^2)^2 + \alpha^2\omega^2}} \sin(\omega t)\right) \\
&= A (\cos\theta\cos(\omega t) + \sin\theta\sin(\omega t)) = A\cos(\omega t-\theta) \tag{8} \end{align}
where we have assigned
\begin{align}
\sin\theta &= \frac{\alpha\omega}{\sqrt{(\omega_0^2-\omega^2)^2 + \alpha^2\omega^2}} & \cos\theta &= \frac{\omega_0^2-\omega^2}{\sqrt{(\omega_0^2-\omega^2)^2 + \alpha^2\omega^2}} \tag{9} \\
A &= \frac{\hat{F}}{m\sqrt{(\omega_0^2-\omega^2)^2 + \alpha^2\omega^2}} \tag{10}
\end{align}
by manipulating trigonometric relationships. \( A \) will be the oscillation amplitude of the forced response and \( \theta \) the phase angle (lagged relative to the input forcing). Usually, the damping \( \alpha \) is a relatively smaller number, so the largest amplitude of \( A \) will occur when \( \omega = \omega_0 \) (where the \( (\omega_0^2-\omega^2)^2 \) part becomes zero):
\begin{align}
A_{\text{max}} &= \frac{\hat{F}}{m\sqrt{\alpha^2\omega_0^2}} \\
&= \frac{\hat{F}}{m\alpha\omega_0} \tag{11}
\end{align}
This is when resonance occurs. This sheds some light on the mechanism behind the deadly collapse of buildings from a severe earthquake or why a seemingly strong bridge can be destroyed by some gusty wind, where the structures have an intrinsic natural frequency \(\omega_0\) similar to the wind frequency \(\omega\).
Approximation Near Resonance
Rewriting (10), we have
\begin{align}
A &= \frac{\hat{F}}{m\sqrt{(\omega_0^2-\omega^2)^2 + \alpha^2\omega^2}} \\
&= \frac{\hat{F}}{m\alpha\omega}\left(\frac{(\omega_0^2-\omega^2)^2}{\alpha^2\omega^2} + 1\right)^{-1/2} \tag{12}
\end{align}
The change in amplitude \( A \) near the resonance point \( \omega \approx \omega_0 \) can be inferred by doing a binomial expansion (\( (1+z)^n \approx 1+nz \)):
\begin{align}
A &\approx \frac{\hat{F}}{m\alpha\omega_0} \left(1 -\frac{1}{2}\frac{(\omega_0^2-\omega^2)^2}{\alpha^2\omega_0^2}\right) \tag{13} \\
\end{align}
(we keep the difference term \( \omega_0^2-\omega^2 \) intact) This shows that the asymptotic behavior of \( A \) at \( \omega \approx \omega_0 \) is a quadratic curve in \(\omega^2\) that peaks at \( \omega = \omega_0 \) (of course) and concave downwards. The width of the peak can be estimated by locating the half-amplitude points:
\begin{align}
A \approx \frac{\hat{F}}{m\alpha\omega_0} \left(1 -\frac{1}{2}\frac{(\omega_0^2-\omega^2)^2} {\alpha^2\omega_0^2}\right) = \frac{1}{2}A_{\text{max}} &= \frac{1}{2}\frac{\hat{F}}{m\alpha\omega_0} \\
1 -\frac{1}{2}\frac{(\omega_0^2-\omega^2)^2}{\alpha^2\omega_0^2} &= \frac{1}{2} \\
(\omega_0^2-\omega^2)^2 &= \alpha^2\omega_0^2 \\
(\omega_0-\omega)^2(\omega_0+\omega)^2 &= \alpha^2\omega_0^2 \\
(\omega_0-\omega)(\omega_0+\omega) &= \pm\alpha\omega_0 \\
(\omega_0-\omega)(2\omega_0) &\approx \pm\alpha\omega_0 \\
\Delta \omega = 2|\omega_0-\omega| &\approx \alpha \tag{14}
\end{align}
So the width of the amplitude peak is of the order \( \alpha \).
Exercise
Find the resonance frequency and amplitude for the RLC circuit system outlined by (12) in this tutorial.
Answer
As before, \( \alpha = R/L \) and \( \omega_0 = \sqrt{1/LC} \). The resonance frequency is exactly \( \omega = \omega_0 = \sqrt{1/LC} \), and if the input potential is sinusoidal \( V(t) = \hat{V}\cos(\omega t) \) (so the forcing term takes the form of \( G(t) = (\hat{V}/L)\cos(\omega t)\)), then replacing by the corresponding variables in (11), we get
\begin{align}
q_{\text{max}} &= \frac{\hat{V}}{L\alpha\omega_0} \\
&= \frac{\hat{V}}{L(R/L)\omega_0} \\
&= \frac{\hat{V}}{R\omega_0}
\end{align}
So the maximum current will be
\begin{align}
I_{\text{max}} &= \left. \frac{dq}{dt} \right|_{\omega = \omega_0 = \sqrt{1/LC}} \\
&= \omega_0\frac{\hat{V}}{R\omega_0} & \begin{aligned}(&\text{The time derivative of}\\ &\text{the sinusoidal wave signal produces \( \omega_0\)})\end{aligned} \\
&= \frac{\hat{V}}{R}
\end{align}
the usual form when there is only the resistor.
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