# 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 friction,
• $\vec{F_f}\,$ represents the friction force and
• $\vec{v}\,$ is the velocity.

Note that the minus sign indicates that the 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:

$m\ddot x\dot{x}+\lambda\dot{x}^2+k x\dot{x}=0$

Now we integrate this equation from 0 to t to obtain an expression for the energy:

$m\frac{\dot{x}^2(t)}{2}+\lambda\frac{x^2(t)}{2}=m\frac{\dot{x}^2(0)}{2}+\lambda\frac{x^2(0)}{2}-\frac{k}{2}\int_{0}^{t}\dot{x}^2dt$

Denoting the mechanical energy by

$E(t):=m\frac{\dot{x}^2(t)}{2}+\lambda\frac{x^2(t)}{2}\,$

the variation of energy is given by:

$E(t)-E(0)=-\frac{k}{2}\int_{0}^{t}\dot{x}^2dt$

That is to say, if 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:

$x(t) = e^{st}\,$

substituting this solution into the equation, we find the quadratic equation:

$m s^2 + \lambda s + k =0\,$

the solution of this equation is given by:

$s=\frac{-\lambda \pm \sqrt{\lambda^2-4mk}}{2m}$