Section 3.7 Harmonic oscillations
Subsection 3.7.1 Mass-spring systems
Suppose a mass m hangs from a vertical spring of original length l\text{.}
\begin{equation*}
mu^{\prime\prime}(t)+\gamma u^{\prime}(t)+ku(t)=F(t)\,\,\,\,\,\,u(0)=u_{0},u^{\prime}(0)=v_{0}.
\end{equation*}
where m,\gamma,k are positive.
The specific constants depend on the measurement system in use. m is found from w=mg\text{.} \gamma is given in units of \frac{\text{weight unit} \cdot s}{\text{distance unit}}\text{.} k is found using Hooke's Law, mg=kL\text{.}
Example 3.7.1.
A \(4\) lb mass stretches a spring \(2\) inches. The mass is displaced an additional 6 in. and then released; and is in a medium that exerts a viscous resistance of \(6\) lb when the mass has a velocity of \(3\) ft/sec. Formulate the IVP that governs the motion of this mass.
Solution:
Find \(m\text{:}\) \(w=mg\) which implies
\begin{equation*}
m=\frac{w}{g}=\frac{4\text{ lb}}{32\text{ ft/s}^2}=\frac{1}{8}\frac{\text{lb}s^{2}}{ft}
\end{equation*}
Find \(\gamma\text{:}\) Using \(\gamma u^{\prime}=6\) lb we have
\begin{equation*}
\gamma=\frac{6\text{ lb}}{3\text{ ft/sec}}=2\frac{\text{lb sec}}{\text{ft}}.
\end{equation*}
Find \(k\text{:}\) (Hooke's Law)
\begin{equation*}
k=\frac{mg}{L}=\frac{4\text{ lb}}{2\text{ in}}=\frac{4\text{ lb}}{(1/6)\text{ ft}}=24\frac{\text{lb}}{ft}.
\end{equation*}
Thus
\begin{equation*}
\frac{1}{8}u^{\prime\prime}+2u^{\prime}+24u=0
\end{equation*}
hence
\begin{equation*}
u^{\prime\prime}+16u^{\prime}+192u=0,\,\,\,\,u(0)=\frac{1}{2},\,\,\,u^{\prime}(0)=0
\end{equation*}
since \(u(0)=6\text{in }\frac{1\text{ft}}{12\text{in}}=\frac{1}{2}\text{.}\)
Solving this
\begin{equation*}
u(t)=\frac{1}{4}e^{-8t}\left(2\cos\left(8\sqrt{2}t\right)+\sqrt{2}\sin\left(8\sqrt{2}t\right)\right).
\end{equation*}
Definition 3.7.2.
When
\begin{equation*}
u(t)=A\cos\omega_{0}t+B\sin\omega_{0}t=R\cos\left(\omega_{0}t-\delta\right)
\end{equation*}
then \omega_{0}= is the natural frequency of the system.
Subsection 3.7.2 Derivation of the mass/spring Differential Equation
To see how the differential equation that models the behavior of a mass-spring system is derived, we will recall some basic facts from Physics. Using Newton's Second Law:
\begin{equation*}
mu^{\prime\prime}(t)=f(t).
\end{equation*}
where f is the net force acting on the mass.
The forces are:
Weight: w=mg (a downward force).
Spring force: F_{s}=-k\left(L+u(t)\right) (either up or down force).
Damping force: F_{d}(t)=-\gamma u^{\prime}(t) (either up or down), which may be due to air resistance, friction, machanical device. It acts in the opposite direction as the motion of the mass.
External force: F(t) (either an up or down).
Putting it all together
\begin{align*}
mu^{\prime\prime}(t) \amp= mg+F_{s}(t)+F_{d}(t)+F(t) \\
\amp= mg-k\left(L+u((t)\right)-\gamma u^{\prime}(t)+F(t)
\end{align*}
which using mg=kL (from Hooke's Law) and simplifying we have
\begin{equation*}
mu^{\prime\prime}(t)+\gamma u^{\prime}(t)+ku(t)=F(t)\,\,\,\,\,\,u(0)=u_{0},u^{\prime}(0)=v_{0}.
\end{equation*}
where m,\gamma,k are positive.Subsection 3.7.3 Undamped harmonic oscillator
When the damping coefficient \gamma=0 (nothing stopping it from oscillating forever), we have
\begin{equation*}
mu^{\prime\prime}+ku=0
\end{equation*}
so that mr^{2}+k=0 gives r=\pm i\sqrt{\frac{k}{m}}.This is a special number, so we'll denote it \omega_{0}=\sqrt{\frac{k}{m}}. We get
\begin{equation*}
u(t)=A\cos\omega_{0}t+B\sin\omega_{0}t
\end{equation*}
with period \frac{2\pi}{\omega} and the natural frequency of the system is \omega_{0}\text{.}Subsection 3.7.4 Damped harmonic oscillator:
A damper is an applied force that resists velocity (that is, the damping force is always opposite the direction of motion). A common way to model a damping force is as proportional to the velocity u'(t)\text{.} When damped, the model equation becomes
\begin{equation*}
mu^{\prime\prime}(t)+\gamma u^{\prime}(t)+ku(t)=0.
\end{equation*}
In general we'll have the following characteristic equation
\begin{equation*}
mr^{2}+\gamma r+k=0,
\end{equation*}
and solving for the roots we get
\begin{equation*}
r=\frac{-\gamma\pm\sqrt{\gamma^{2}-4km}}{2m}.
\end{equation*}
Subsection 3.7.5 Types of oscillating systems
Different types of behavior are possible depending on the value of b^{2}-4mk\text{.} We'll classify the possible cases in the following way:-
If \gamma=0\text{,}
- the oscillator is undamped.
- Mass oscillates forever
- The natural period is 2\pi\sqrt{\frac{m}{k}}\text{.}
-
If \gamma>0 and \gamma^{2}-4km\lt0 (which happens when the roots are r=\alpha\pm\beta i)
- The oscillator is underdamped. The mass oscillates back and forth as it tends to its rest position. The solutions are\begin{equation*} u=Re^{-\gamma t/(2m)}\cos\left(\mu t-\delta\right) \end{equation*}and u is bounded between \pm Re^{-\gamma t/(2m)}\text{.}
- The oscillator is underdamped. The mass oscillates back and forth as it tends to its rest position. The solutions are
-
If \gamma>0 and \gamma^{2}-4km>0 (which happens when there are two distinct r_{1},r_{2}):
- The oscillator is overdamped. The mass tends to its rest position but does not oscillate.
- The solutions are\begin{equation*} u=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t},\,\,\,r_{1},r_{2}\lt0 \end{equation*}
-
If \gamma>0 and \gamma^{2}-4km=0 (which happens when there is one negative r):
- The oscillator is critically damped. The mass tends to its rest position but does not oscillate.
- Solutions tend to the origin tangent to the unique line of eigenvectors.
- The solutions are\begin{equation*} u=c_{1}e^{-\gamma t/(2m)}+c_{2}te^{-\gamma t/(2m)} \end{equation*}
Subsection 3.7.6 Electric Circuts
The flow of electric charge in certain basic electrical circuits (RLC for resistor (R), inductor (L), capacitor (C)) is modeled by second order linear ODEs with constant coefficients:
\begin{equation*}
LQ^{\prime\prime}(t)+RQ^{\prime}(t)+\frac{1}{C}Q(t)=E(t),\,\,\,\,\,Q(0)=Q_{0},\,\,\,Q^{\prime}(0)=Q_{0}^{\prime}
\end{equation*}
where Q= charge (coulombs).
