---
# System prepended metadata

title: 2020.4.15(10)
tags: [Mechanics, Osciliation]

---

---
tags : Mechanics,Osciliation
title : 2020.4.15(10)
---
# Mechanics (10)
## Damped osciliation
Think about an object with a spring (spring constant : k) faced with viscous drag $-2hv$

$\large F=-kx-2hv$
$\large\Leftrightarrow m\frac{d^2x}{dt^2} = -kx-2hv$

Divide with $m$ , and transfer right to left

$\large \ddot{x}+2\gamma\dot{x}+\omega _{0}^2 x =0$ 

$(\because \gamma \equiv \frac{h}{m} , \omega_0 ^2 \equiv \frac{k}{m} )$

Solve this defferentional equation

Set as $x=e^{-\gamma t}f(t)$ ,

$\large (\ddot{f}-2\gamma\dot{f}+\gamma^2f+2\gamma(\dot{f}-\gamma f)+\omega_0 ^2 f)e^{-\gamma t} = 0$

$\large \Leftrightarrow (\ddot{f}+(\omega_0 ^2 -\gamma ^2)f)e^{-\gamma t} = 0$

Therefore, we can get $\ddot{f}+(\omega_0^2 -\gamma^2)f = 0$ .

Think as 3patern about the solution of this equation.

### 1~st~ Damped Osciliation
 When viscous drag is low and fit $\gamma < \omega_{0}$ , define $\omega_1 = \sqrt{\omega_0 ^2 -\gamma^2} ,$ for $\ddot{f}+\omega_1 ^2 f =0 ,$ for the sake of setting the solution as $f=acos(\omega_1 t+ \delta)$ , we can get 
 
 $\huge x(t)=ae^{-\gamma t}cos(\omega_1t+\delta)$
 
 Ampitude decrease exponentially in relaxation time $\frac{1}{\gamma}$.
 $\omega_1$ decrease rather than $\omega_0$ , so the period $T = \frac{2\pi}{\omega_1}$ become longer.
 ![](https://i.imgur.com/jRWfYBb.jpg)
