starting oscilliations

content/physics/harmonic-motion.md

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+---
+title: Less Simple Harmonic Motion
+date: 2025-01-09
+---
+
+# Simple harmonic motion - Spring oscillations
+
+When an object *vibrates* or *oscillates* back and forth, over the same path, each oscillation taking the same amount of time, the motion is **periodic.** The simplest form of periodic motion closely resemble this system, we will examine ... blah blah blah. Any spring at natural length is defined to be at its **equilibrium position.** If additional energy is imparted upon the system, either compressing or stretching said spring, a restoring force will occur to *restore* the initial position.
+
+$$
+F=\pm kx
+$$
+
+Terms, I guess:
+
+- the distance from the equilibrium point is *displacement*
+- Peak displacement is called the *amplitude*
+- one *cycle* is the distance covered between the trough of the position curve over time and the peak of the same curve
+- the *period* $T$ is the time required to complete one cycle's worth of motion 
+- the *frequency* $f$ is the number of complete cycles per second. 
+
+$$
+f=\frac{1}{T} \text{ and } T=\frac{1}{f}
+$$
+
+# Energy in Simple Harmonic Motion 
+
+Please just go learn Lagrangian Mechanics already. The energy approach is obviously the most useful of them all. Elastic Potential energy:
+
+$$
+|\int{F\mathrm{d}x}|=\int{kx \mathrm{d}x}=\tfrac{1}{2}kx^2 
+$$
+
+The total energy: $E=\textbf{KE}+\textbf{PE}$
+
+$$
+E=\tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 
+$$
+
+We all know what $x$ and $v$ are. SHM can only occur if friction and other external forces are approximately net zero. $A$ is defined as the absolute peak distance from equilibrium. When $x=\pm A$, the $\textbf{KE}=0$.
+
+$$
+\tfrac{1}{2}m0^2+\tfrac{1}{2}kA^2 
+$$
+
+At equilibrium, velocity and therefore kinetic energy are the greatest, with $\textbf{PE}=0$:
+
+$$
+E=\tfrac{1}{2}m(v_{\text{max}})^2 
+$$
+
+Since energy is always conserved in SHM:
+
+$$
+\tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 = \tfrac{1}{2}kA^2 
+$$
+
+Solving the system of equations for $v$:
+
+$$
+v^2 = \frac{k}{m}A^2(1-\frac{x^2}{A^2})
+$$
+Given $mv^2=kA^2$:
+$$
+v_max=\sqrt{\frac{k}{m}}A 
+$$
+
+With another step of substitution:
+
+$$
+v=\pm v_{\text{max}}\sqrt{1-\frac{x^2}{A^2}}
+$$
+
+# The Period and the Sinusoidal Nature of SHM
+
+The period of harmonic motion depends on the stiffness of the spring and the mass $m$ in question. It does not, however, depend on the amplitude ,$A$.
+
+$$
+T=2\pi\sqrt{\tfrac{m}{k}}
+$$
+
+Larger mass means more rotational inertial (we haven't gotten to this yet), so that part of it makes sense. It is also important that $k\propto F$
+
+$$
+f=\frac{1}{t}=\frac{1}{2\pi}\sqrt{\frac{k}{m}}
+$$
+