Question

Establish the differential equation for damped harmonic oscillator and obtain its solution.
Show that the damped oscillator will exhibit non-oscillatory behaviour if the damping is
heavy.

Answer 1

Answer on Question #40591, Physics, Mechanics | Kinematics | Dynamics

Establish the differential equation for damped harmonic oscillator and obtain its solution. Show that the damped oscillator will exhibit non-oscillatory behavior if the damping is heavy.

Solution

An ideal mass-spring-damper system with mass $m$, spring constant $k$ and viscous damper of damping coefficient $c$ is subject to an oscillatory force

$$F_{\mathrm{osc}} = -kx,$$

and a damping force

$$F_{\mathrm{d}} = -cv = -c \frac{dx}{dt}.$$

Applying Newton's second law, the total force $F$ on the body is

$$F = ma = m \frac{d^2x}{dt^2},$$

where $a$ is the acceleration of the mass and $x$ is the displacement of the mass relative to a fixed point of reference.

Since $F = F_{\mathrm{osc}} + F_d$

$$m \frac{d^2x}{dt^2} = -kx - c \frac{dx}{dt}.$$

This differential equation may be rearranged into

$$\frac{d^2x}{dt^2} + \frac{c}{m} \frac{dx}{dt} + \frac{k}{m}x = 0 \text{ or } \left\{\frac{d^2}{dt^2} + \frac{c}{m} \frac{d}{dt} + \frac{k}{m}\right\} x = 0.$$

In this differential equation $\frac{d}{dt} = D$ is differential operator. So

$$\left\{D^2 + \frac{c}{m} D + \frac{k}{m}\right\} x = 0.$$

Let's complete the square

$$\left\{D^2 + \frac{c}{m} D + \frac{k}{m}\right\} x = \left\{\left(D + \frac{c}{2m}\right)^2 + \left(\frac{k}{m} - \left(\frac{c}{2m}\right)^2\right)\right\} x = 0.$$

We can set

$$\omega^2 = \frac{k}{m} - \left(\frac{c}{2m}\right)^2.$$

We are assuming here that $\omega^2 \geq 0$.

Now our differential equation reads

$$\left\{\left(\frac{d}{dt} + \frac{c}{2m}\right)^2 + \omega^2\right\} x = 0 \text{ or } \left(\frac{d}{dt} + \frac{c}{2m}\right)^2 x = -\omega^2 x.$$

Let's take the square root of both sides of this equation

$$\left(\frac {d}{d t} + \frac {c}{2 m}\right) x = \pm i \omega x.$$

Now we have two first order equations to solve:

$$\frac {d x}{d t} = \left(- \frac {c}{2 m} \pm i \omega\right) x,$$

which have the solution

$$x = e ^ {- \frac {c}{2 m} t} (A \sin \omega t + B \cos \omega t).$$

It is the general form of the solution representing damped oscillations with damped frequency

$$\omega = \sqrt {\frac {k}{m} - \left(\frac {c}{2 m}\right) ^ {2}}.$$

The overdamped case (damping is heavy) occurs when $\omega_0 = \frac{k}{m} < \frac{c}{2m}$. Now the system doesn't oscillate at all; the motion simply dies away. This is characterised by a solution which decays exponentially.

Let's rewrite our equation

$$\left\{\left(\frac {d}{d t} + \frac {c}{2 m}\right) ^ {2} - \omega^ {2} \right\} x = 0$$

where $\omega^2 = \left(\frac{c}{2m}\right)^2 -\frac{k}{m} >0.$

Then upon square rooting our equation we obtain

$$\left(\frac {d}{d t} + \frac {c}{2 m}\right) x = \pm \omega x,$$

which have a solution

$$x = C e ^ {(- \frac {c}{2 m} + \omega) t} + D e ^ {(- \frac {c}{2 m} - \omega) t}.$$

This solution represents a damped motion without oscillations.