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

Note that the minus sign indicates that the damping friction force always opposes the movement.
 * $$\lambda\,$$ 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.

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: $$m{{{{\dot x}^2}(t)} \over 2} + k{{{x^2}(t)} \over 2} = m{{{{\dot x}^2}(0)} \over 2} + k{{{x^2}(0)} \over 2} - {\lambda \over 2}\int_0^t  dt$$ 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: $$E(t) - E(0) = - {\lambda  \over 2}\int_0^t  dt$$ 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. 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 $$x<0$$.
 * 1) The starting position of the mass. $$x_2$$
 * 2) The starting direction and magnitude of motion. $$v$$

For upward motion $$v<0$$, and for downward motion $$v>0$$.

Solution
We look for a general solution in the following form: $$x(t) = A_1e^{s_1t} + A_2e^{s_2t}\,$$ 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: $$s_1,s_2=\frac{-\lambda \pm \sqrt{\lambda^2-4mk}}{2m}$$ 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.