# Ordinary 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.

1. The starting position of the mass. $x_{2}$
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.