Differential Equations/Motion with a Damping Force

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Differential Equations > Applications of Second-Order Differential Equations > Motion with a Damping Force

Contents

[edit] Introduction

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 influency of friction.

[edit] 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.


[edit] 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

[edit] 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.


[edit] 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. x2
  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.

[edit] 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}

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