Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$x''+\gamma x'+w_0^2x=0$

That is the general equation for damped harmonic motion. What is the term or name that describes gamma ?

Is it called the damping constant ? I know its the ration between the resistive coefficient (b) and mass of the system (m) but what do we actually call it ?

share|cite|improve this question
I changed your tag from "harmonic functions" to "differential equations." Harmonic functions are something completely different. – icurays1 Apr 28 '13 at 19:07
up vote 0 down vote accepted

The second order differential equation arises from the application of Newton's Second Law.

$$\sum F = ma$$ In the case of oscillatory systems, such as a spring, there are two forces exerted onto the spring. The restoring force $-kx$ and the damping force $-bv$ where $v$ is the velocity of the spring. Note that the assumption of linear resistive force is only an approximation, and at higher velocities drag is actually proportional to the square of velocity.

$$\begin{align} -kx-bv &= ma\\ m\ddot{x} + b\dot{x}+kx&=0\\ \ddot{x}+\frac{b}{m}\dot{x}+\frac{k}{m}x &= 0 \end{align}$$ which is the equation you had above.

share|cite|improve this answer

It's called damping ratio, damping coefficient of damping constant. it measures how the oscillations of the system decay after an initial force is applied. You can calculate it with the expression: $$\gamma=\frac{c}{\sqrt{km}}$$ where $c$ is the friction coefficient, $m$ the mass of the oscillating object and $k$ the elastic constant corresponding to Hooke's law. If $\gamma>1$ we say that the oscillator is overdamped.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.