Simple harmonic oscillator

The simple harmonic oscillator

Simple harmonic motion

For SHM to occur we require stable equilibrium, about a point. For example, at the origin we could have: $$\sum \vec{F}(0)=0\text{,}$$ which would describe a system in equilibrium. This however is not necessarily stableequilibrium.

The lower part of the figure shows the case of unstable equilibrium. The upper part shows the case of stable equilibrium. These situations often occur inmechanical systems.

For example, consider a mass attached to a spring:

In general, in a case of stable equilibrium we can write the force as a polynomial expansion: $$F(x)=-(k_{1}x+k_2{x}^2+k_3{x}^3+\dots )$$ where the $k_i$ are positive constants. There is always a region of $x$ small enough that we can write: $$F=-kx$$ $${\displaystyle \begin{array}{c}F=-kx\\ ma=-kx\\ m\ddot{x}=-kx\\ \ddot{x}+\frac{k}{m}x=0\end{array}}$$ This is satisfied by an equation of the form $$x=A\sin (\omega t+{\phi }_0)$$ where $A$ and ${\phi }_0$ are constants that are determined by the initial conditions. Draw a diagram of a sinusoid and mark on it the period T and Amplitude A

${\phi }_0$ Is an arbitrary phase which shifts the sinusoid.This is also satisfied by an equation of the form $$x=A\sin (\omega t)+B\cos (\omega t)$$ Lets show this: $${\displaystyle \begin{array}{c}x=A\sin (\omega t)+B\cos (\omega t)\\ \dot{x}=\omega (A\cos (\omega t)-B\sin (\omega t))\\ \ddot{x}=-{\omega }^2(A\sin (\omega t)+B\cos (\omega t))\\ \ddot{x}=-{\omega }^{2}x\end{array}}$$ Again there are two constants determined by the initial conditions $A$ and $B$ The equation can be rewritten $$\ddot{x}+{\omega }^{2}x=0$$ Thus if $${\omega }^2=\frac{k}{m}$$ then the equation is identical to the SHM equation.

So another way to write the equation of Simple Harmonic Motion is $$\ddot{x}+{\omega }^{2}x=0$$ or $$\ddot{x}=-{\omega }^{2}x$$

It is also important to remember the relationships between freqency, angular frequency and period: $${\displaystyle \begin{array}{c}\omega =2\pi \nu \\ T=\frac{2\pi }{\omega }\\ \nu =\frac{1}{T}\end{array}}$$

Another solution to the SHM equation is $$\tilde{x}=A\cos (\omega t+{\phi }_0)+iA\sin (\omega t+{\phi }_0)$$ Recall Taylor's expansions of sine and cosine $$\sin \theta =\theta -\frac{{\theta }^3}{3!}+\frac{{\theta }^5}{5!}\dots$$ $$\cos \theta =1-\frac{{\theta }^2}{2!}+\frac{{\theta }^4}{4!}\dots$$ Then $${\displaystyle \begin{array}{c}\cos \theta +i\sin \theta =1+i\theta -\frac{{\theta }^2}{2!}-i\frac{{\theta }^3}{3!}+\frac{{\theta }^4}{4!}\dots \\ =1+i\theta +\frac{{(i\theta )}^2}{2!}+\frac{{(i\theta )}^3}{3!}+\frac{{(i\theta )}^4}{4!}\dots \\ =e^{i\theta }\end{array}}$$

(an alternative way to show this is the following) $${\displaystyle \begin{array}{c}z\equiv \cos \theta +i\sin \theta \\ dz=(-\sin \theta +i\cos \theta )d\theta =izd\theta \\ \int \frac{dz}{z}=\int id\theta \\ \ln z=i\theta \\ z=e^{i\theta }\end{array}}$$

Thus we can write $$\tilde{x}=A\cos (\omega t+{\phi }_0)+iA\sin (\omega t+{\phi }_0)$$ as $$\tilde{x}=A{e}^{i(\omega t+{\phi }_0)}$$ $${\displaystyle \begin{array}{c}\tilde{x}=A{e}^{i(\omega t+{\phi }_0)}\\ \dot{\tilde{x}}=i\omega A{e}^{i(\omega t+{\phi }_0)}\\ \ddot{\tilde{x}}={(i\omega )}^{2}A{e}^{i(\omega t+{\phi }_0)}=-{\omega }^2\tilde{x}\end{array}}$$