# Solutions to the Damped Oscillator Equation

Author

jithpadduri.123

Last Updated

5년 전

License

Creative Commons CC BY 4.0

Abstract

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

Author

jithpadduri.123

Last Updated

5년 전

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}
```

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