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 #44795, Physics, Mechanics | Kinematics | Dynamics

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:

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:

$$F_s = -kx$$

and a damping force

$$F_d = -cv = -c \frac{dx}{dt} = -c\dot{x}$$

Treating the mass as a free body and applying Newton's second law, the total force $F_{tot}$ on the body is:

$$F_{tot} = -kx - c\dot{x} = m\ddot{x}$$

This differential equation may be rearranged into

$$\ddot{x} + \frac{c}{m}\dot{x} + \frac{k}{m}x = 0$$

The following parameters are then defined:

$$\omega_0 = \sqrt{\frac{k}{m}}, \zeta = \frac{c}{2\sqrt{mk}}$$ $$\ddot{x} + 2\zeta\omega_0\dot{x} + \omega_0^2x = 0$$

Continuing, we can solve the equation by assuming a solution $x$ such that:

$$x = e^{\gamma t}$$

where the parameter $\gamma$ is, in general, a complex number.

Substituting this assumed solution back into the differential equation gives:

$$\gamma^{2} + 2 \zeta \omega_{0} \gamma + \omega_{0}^{2} x = 0$$

Solving the characteristic equation will give two roots:

$$\gamma_{\pm} = - \zeta \omega_{0} \pm \sqrt{(\zeta \omega_{0})^{2} - \omega_{0}^{4}}$$

The solution to the differential equation is thus:

$$x(t) = A e^{\gamma_{+} t} + B e^{\gamma_{-} t}$$

where A and B are determined by the initial conditions of the system:

$$A = x(0) + \frac{\gamma_{+} x(0) - \dot{x}(0)}{\gamma_{-} - \gamma_{+}}$$ $$B = - \frac{\gamma_{+} x(0) - \dot{x}(0)}{\gamma_{-} - \gamma_{+}}$$

When $\zeta > 1$, the system is over-damped and there are two different real roots.

The solution to the motion equation is:

$$x(t) = A e^{\gamma_{+} t} + B e^{\gamma_{-} t}$$

with real $\gamma_{+}$ and $\gamma_{-}$ therefore it is non-oscillatory behavior.

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