Deriving the equations of motion for this model is usually done by summing the forces on the mass (including any applied external forces ${\displaystyle F_{\text{external}})}$ :

${\displaystyle \Sigma F=-kx-c{\dot {x}}+F_{\text{external}}=m{\ddot {x}}}$ 

By rearranging this equation, we can derive the standard form:

${\displaystyle {\ddot {x}}+2\zeta \omega _{n}{\dot {x}}+\omega _{n}^{2}x=u}$  where ${\displaystyle \omega _{n}={\sqrt {\frac {k}{m}}};\quad \zeta ={\frac {c}{2m\omega _{n}}};\quad u={\frac {F_{\text{external}}}{m}}}$ 

${\displaystyle \omega _{n}}$  is the undamped natural frequency and ${\displaystyle \zeta }$  is the damping ratio. The homogeneous equation for the mass spring system is:

${\displaystyle {\ddot {x}}+2\zeta \omega _{n}{\dot {x}}+\omega _{n}^{2}x=0}$ 

This has the solution:

${\displaystyle x=Ae^{-\omega _{n}t\left(\zeta +{\sqrt {\zeta ^{2}-1}}\right)}+Be^{-\omega _{n}t\left(\zeta -{\sqrt {\zeta ^{2}-1}}\right)}}$ 

If ${\displaystyle \zeta <1}$  then ${\displaystyle \zeta ^{2}-1}$  is negative, meaning the square root will be imaginary and therefore the solution will have an oscillatory component.^[2]

• Numerical methods
• Soft body dynamics#Spring/mass models
• Finite element analysis