# Fundamental Physics/Motion/Oscillation/Spring Oscillations

## Spring's Oscillations

Force mass to stretch out vertically

${\displaystyle F_{a}=ma}$

${\displaystyle F_{y}=-ky}$

the system is at equilibrium when the sum of forces act on the mass equals to zero

${\displaystyle F_{a}+F_{y}=0}$
${\displaystyle ma=-ky}$
${\displaystyle a=-{\frac {k}{m}}y}$

At all time

${\displaystyle a(t)=-{\frac {k}{m}}y(t)}$
${\displaystyle {\frac {d^{2}}{dt^{2}}}y(t)=-{\frac {k}{m}}y(t)}$
${\displaystyle s^{2}y(t)=-{\frac {k}{m}}y(t)}$
${\displaystyle s^{2}=-{\frac {k}{m}}}$
${\displaystyle s=\pm j{\sqrt {\frac {k}{m}}}}$
${\displaystyle y(t)=Ae^{st}=Ae^{\pm j{\sqrt {\frac {k}{m}}}t}=ASin\omega t}$

## Summary

Spring's oscillation can be characterized by an Oscillation equation, a second ordered differential equation of the form which has a root called Wave function

${\displaystyle {\frac {d^{2}}{dt^{2}}}y(t)=-{\frac {k}{m}}y(t)}$
${\displaystyle y(t)=ASin\omega t}$