# Ordinary Differential Equations/Motion with a Damping Force

Applications of Second-Order Differential Equations > Motion with a Damping Force
Simple Harmonic Motion with a Damping Force can be used to describe the motion of a mass at the end of a spring under the influence of friction.

## Laws of Motion

The friction force is considered to obey a linear law, that to say, it is given by the following expression:
    ${\vec {F_{f}}}=-\lambda {\vec {v}}\,$ where
* $\lambda \,$ is a positive constant and represents the coefficient of damping friction force,
* ${\vec {F_{f}}}\,$ represents the friction force and
* ${\vec {v}}\,$ is the velocity.
Note that the minus sign indicates that the damping friction force always opposes the movement.

## The Differential Equation of the Motion

The differential equation of the motion with a damping force will be given by:
$m{\ddot {x}}+\lambda {\dot {x}}+kx=0$
In order to obtain the leading coefficient equal to 1, we divide this equation by the mass:
${\ddot {x}}+{\frac {\lambda }{m}}{\dot {x}}+{\frac {k}{m}}x=0$

### Non-conservation of energy

We may multiply the equation of motion by the velocity ${\dot {x}}\,$ in order to get an integrable form:
$m{\ddot {x}}{\dot {x}}+\lambda {\dot {x}}^{2}+kx{\dot {x}}=0$
Now we integrate this equation from 0 to t to obtain an expression for the energy:
$m{{{{\dot {x}}^{2}}(t)} \over 2}+k{{{x^{2}}(t)} \over 2}=m{{{{\dot {x}}^{2}}(0)} \over 2}+k{{{x^{2}}(0)} \over 2}-{\lambda \over 2}\int _{0}^{t}{{\dot {x}}^{2}}dt$
Denoting the mechanical energy by
$E(t):=m{{{{\dot {x}}^{2}}(t)} \over 2}+k{{{x^{2}}(t)} \over 2}\,$
the variation of energy is given by:
$E(t)-E(0)=-{\lambda \over 2}\int _{0}^{t}{{\dot {x}}^{2}}dt$
That is to say, if the damping friction force coefficient $\lambda$ is not zero, or integration over square of the velocity does not vanish, the system is losing energy. Physically speaking, friction converts mechanical energy into thermal energy.

### Initial condition

With the free motion equation, there are generally two bits of information one must have to appropriately describe the mass's motion.
* The starting position of the mass. $x_{2}$
* The starting direction and magnitude of motion. $v$
Generally, one isn't present without the other. For simplicity, we will consider all displacement below the equilibrium point as $x>0$ and above as $x<0$.
For upward motion $v<0$, and for downward motion $v>0$.

### Solution

We look for a general solution in the following form:
$x(t)=A_{1}e^{s_{1}t}+A_{2}e^{s_{2}t}\,$
substituting this solution into the equation, we find the quadratic equation:
$ms^{2}+\lambda s+k=0\,$
the solution of this equation is given by:
$s_{1},s_{2}={\frac {-\lambda \pm {\sqrt {\lambda ^{2}-4mk}}}{2m}}$
And $A_{1},A_{2}$ are determined by initial conditions. Obviously, this solution may have real-valued or complex-valued roots. In any case, the real part of the roots is always negative (since both $k$ and $m$ are positive), implying stable solution. When both roots are real-valued, the system is called over-damped, whilst it has two complex roots (where one is the complex conjugate of the other) the system is called under-damped. In case $\lambda$=0, both roots has zero real-parts, and the solution is oscillating, i.e energy-conserving.