Mars Ultor
4744fd878c
Some checks failed
Build and Test / build-and-test (macos-latest) (push) Has been cancelled
Build and Test / build-and-test (ubuntu-latest) (push) Has been cancelled
Build and Test / build-and-test (windows-latest) (push) Has been cancelled
Build and Test / publish-tag (push) Has been cancelled
88 lines
2.5 KiB
Markdown
88 lines
2.5 KiB
Markdown
---
|
|
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 $k\propto F$
|
|
|
|
$$
|
|
f=\frac{1}{t}=\frac{1}{2\pi}\sqrt{\frac{k}{m}}
|
|
$$
|
|
|