The Laplace Transform: Forced Vibration without Damping and Resonance

Forced Vibration without Damping

Let us consider forced harmonic oscillation without damping
$$m\ddot{X}(t)+kX(t)=F(t)$$
with initial conditions $X(0)=x_0$ and $\dot{X}(0)=v_0$. The transformed equation is
$$m[s^2x(s)-sx_0-v_0]+kx(s)=f(s)$$
from which we obtain
\begin{equation}\begin{aligned}
x(s)&=\frac{sx_0+v_0}{s^2+\omega_0^2}+\frac{1}{m}\frac{f(s)}{s^2+\omega_0^2}\\&=\frac{s}{s^2+\omega_0^2}x_0+\frac{v_0}{\omega_0}\frac{\omega_0}{s^2+\omega_0^2}+\frac{f(s)}{m\omega_0}\frac{\omega_0}{s^2+\omega_0^2}
\end{aligned}\label{eq:laplace18}\end{equation}
where $\omega_0=\sqrt{\frac{k}{m}}$. By taking the inverse transform of \eqref{eq:laplace18} we find the solution $X(t)$ as
\begin{equation} \begin{aligned}
X(s)&=x_0\cos\omega_0t+\frac{v_0}{\omega_0}\sin\omega_0t+\frac{1}{m\omega_0}F(t)\ast\cos\omega_0t\\
&=x_0\cos\omega_0t+\frac{v_0}{\omega_0}\sin\omega_0t+\frac{1}{m\omega_0}\int_0^t\sin\omega_0(t-\tau)F(\tau)d\tau \end{aligned}\label{eq:laplace19}\end{equation}

Resonance

Let $F(t)$ be the sinusoidal force $F(t)=F_0\sin\omega t$ where $F_0$ and $\omega$ are positive constants. Then
$$f(s)=\mathcal{L}{F(t)}=F_0\frac{\omega}{s^2+\omega^2}$$
and \eqref{eq:laplace18} can be written
$$x(s)=\frac{x_0s+v_0}{s^2+\omega_0^2}+\frac{F_0}{m}\frac{\omega}{(s^2+\omega_0^2)(s^2+\omega^2)}$$
If $\omega\ne\omega_0$, $\frac{1}{(s^2+\omega_0^2)(s^2+\omega^2)}=\frac{1}{\omega^2-\omega_0^2}\left[\frac{1}{s^2+\omega_0^2}-\frac{1}{s^2+\omega^2}\right]$ and
$$x(s)=\frac{s}{s^2+\omega_0^2}x_0+\frac{v_0}{\omega_0}\frac{\omega_0}{s^2+\omega_0^2}+\frac{F_0\omega}{m\omega_0(\omega^2-\omega_0^2)}\frac{\omega_0}{s^2+\omega_0^2}-\frac{F_0}{m(\omega^2-\omega_0^2)}\frac{\omega}{s^2+\omega^2}$$
Thus by taking the inverse transform we obtain $X(t)$
$$X(t)=x_0\cos\omega_0t+\frac{1}{\omega_0}\left[v_0+\frac{F_0\omega}{m(\omega^2-\omega_0^2)}\right]\sin\omega_0t-\frac{F_0}{m(\omega^2-\omega_0^2)}\sin\omega t$$
The motion is the superposition of two simple harmonic motions, one with frequency $\omega_0$ (the natural component of the vibration) and the other with frequency $\omega$ (the forced component of the vibration). The natural vibrations are not present if
$$x_0=0,\ v_0=\frac{F_0\omega}{m(\omega_0^2-\omega^2)}$$

When $\omega=\omega_0$,
\begin{align*}x(s)&=\frac{x_0s+v_0}{s^2+\omega_0^2}+\frac{F_0}{m}\frac{\omega_0}{(s^2+\omega_0^2)^2}\\&=\frac{s}{s^2+\omega_0^2}x_0+\frac{v_0}{\omega_0}\frac{\omega_0}{s^2+\omega_0^2}+\frac{F_0}{2m}\frac{1}{s}\frac{2\omega_0s}{s^2+\omega^2} \end{align*}
In the first example here, we found $\mathcal{L}^{-1}\left\{\frac{2\omega_0s}{(s^2+\omega_0^2)^2}\right\}=t\sin\omega_0t$. Using the formula here
\begin{align*} X(t)&=x_0\cos\omega_0t+\frac{v_0}{\omega_0}\sin\omega_0t+\frac{F_0}{2m}\int_0^t\tau\sin\omega_0\tau d\tau\\&=x_0\cos\omega_0t+\frac{1}{\omega_0^2}\left(v_0\omega_0+\frac{F_0}{2m}\right)\sin\omega_0t-\frac{F_0}{2m\omega_0}t\cos\omega_0t \end{align*}
In view of the last term, the amplitude of the oscillations increases indefinitely. In this case, the force $F(t)$ is said to be in resonance with the system.

Vibrations can be quite destructive. A striking example is the collapse of Tacoma Narrows Bridge in 1940. In my college mechanics class I was told that this was resulted from resonance. However the catastrophic vibrations were not actually due to simple mechanical resonance but to a complicated interaction between the bridge and the winds passing through it. Such a phenomenon is called flutter, which is a kind of self-oscillation.

The following is a footage of the old Tacoma Narrows Bridge collapsing: