# Solutions to the Damped Oscillator Equation

Author

jithpadduri.123

License

Creative Commons CC BY 4.0

Abstract

This paper describes the three conditions of damped oscillations and the mathematical formulations of each condition.

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Author

jithpadduri.123

License

Creative Commons CC BY 4.0

Abstract

This paper describes the three conditions of damped oscillations and the mathematical formulations of each condition.

```
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amssymb}
\title{Solutions to the Damped Oscillator Equation}
\author{}
\date{September 2019}
\begin{document}
\maketitle
\noindent The Damped Oscillator Equation \[m\frac{d^2x}{dt^2}+\beta\frac{dx}{dt}+kx=0\] is a second-order differential equation that can easily be derived using Newton's Second Law of Motions applied to a spring, assuming a frictional force $f_k=-\beta v$, where $v$ is the speed of the mass. Solving this equation, however, is considerably more difficult, but not that hard if proper techniques are used.
\section{Over-damped Oscillators}
An over-damped oscillator is an oscillator that faces a frictional force so great that the oscillator just returns to its equilibrium position. Looking at the equation \[m\frac{d^2x}{dt^2}+\beta\frac{dx}{dt}+kx=0\] It can be proved that if $y_1$ and $y_2$ are two linearly independent solutions, meaning that one cannot be multiplied by a number to get the other, then $y=c_1y_1+c_2y_2$ must be a solution. Looking at properties of functions, $e^{rx}$ seems like a good solution to the equation. If $e^{rx}$ is indeed a solution, then plugging back into the main equation gives \[mr^2e^{rx}+\beta re^{rx}+ke^{rx}=(mr^2+\beta r +k)e^{rx}=0\] That means that if $e^{rx}$ is a solution $mr^2+\beta r+k=0$. This is the auxiliary equation, and notice that we can get two roots, one root, or zero real roots. An over-damped oscillator will result in two roots in the equation. If $r_1$ and $r_2$ are solutions, then from the theorem above, the solution must be \[x(t)=c_1e^{r_1x}+c_2e^{r_2x}\] These two roots are \[r=\frac{-\beta\pm\sqrt{\beta^2-4mk}}{2m}\] Thus, \[x(t)=c_1e^{\frac{-\beta+\sqrt{\beta^2-4mk}}{2m}t}+c_2e^{\frac{-\beta-\sqrt{\beta^2-4mk}}{2m}t}\] The constants $c_1$ and $c_2$ can be determined by the amplitude and the initial velocity of the oscillation.
\section{Critically Damped Oscillators}
A critically damped oscillator just slightly overshoots the equilibrium position and gently returns to equilibrium. In this case, the auxiliary equation derived in section 1 \[mr^2+\beta r+k=0\] must have one solution. Let $r$ be the one solution. Then \[r=-\frac{\beta}{2m}.\] However, we cannot use the solution above by putting $r$ for both $r_1$ and $r_2$ as that would result in two linearly dependent terms. However, if we know that $e^{rx}$ is a solution, then it can be verified that $y=xe^{rx}$ is a solution using substitution into the original equation and the fact that $2mr+\beta=0$. Thus, a solution is \[x(t)=c_1e^{-\frac{\beta}{2m}t}+c_2te^{-\frac{\beta}{2m}t}=(c_1+c_2t)e^{-\frac{\beta}{2m}t}\]
\section{Under-damped Oscillators}
An under-damped oscillator will oscillate, but the amplitude will gradually decrease and the oscillations will gradually die out. The auxiliary equation \[mr^2+\beta r+k=0\] has no real solutions but two \textit{complex} solutions. Let $r_1$ and $r_2$ be solutions. Let $r_1=a+bi$ and $r_2=a-bi$. Using the general solution from section 1, we can show that \[x(t)=C_1e^{(a+bi)t}+C_2e^{(a-bi)t}=e^{at}(C_1e^{ibt}+C_2e^{-ibt})\] Using Euler's Formula $e^{i\theta}=\cos\theta+i\sin\theta$, we can show that \[x(t)=e^{at}(C_1(\cos{bt}+i\sin{bt})+C_2(\cos{bt}-i\sin{bt})\] Simplifying, we get \[x(t)=e^{at}[(C_1+C_2)\cos{bt}+i(C_1-C_2)\sin{bt}]=e^{at}(c_1\cos{bt}+c_2\sin{bt})\] where $c_1=C_1+C_2$ and $c_2=i(C_1-C_2)$. If we are looking for real solutions, then $c_1$ and $c_2$ must be real. Back to the original equation, we get \[r=-\frac{\beta}{2m}\pm\frac{i\sqrt{4mk-\beta^2}}{2m}\]Thus, the solution is \[x(t)=e^{-\frac{\beta}{2m}t}\bigg[c_1\cos\bigg(\frac{\sqrt{4mk-\beta^2}}{2m}t\bigg)+c_2\sin\bigg(\frac{\sqrt{4mk-\beta^2}}{2m}t\bigg)\bigg]\]
\end{document}
```