Ordinary Differential Equations/Motion with a Damping Force

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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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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}