Question

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.

Answer 1

Answer on Question #39278, 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:

Newton's Law for a spring system with linear damping reads

$$- k x - b v = m a$$

for a block of mass $m$ attached to a spring of constant $k$ with damping coefficient $b$ .

Figure: Plots of displacement vs time for the mass-spring system: (a) underdamped – mass in air; (b) overdamped – mass in thick oil; (c) critically damped – mass in water

Using the definitions of velocity and acceleration we can write this as the differential equation

$$\frac {d ^ {2} x}{d t ^ {2}} + \frac {b}{m} \frac {d x}{d t} + \frac {k}{m} x = 0$$ $$\left\{\frac {d ^ {2}}{d t ^ {2}} + \frac {b}{m} \frac {d}{d t} + \frac {k}{m} \right\} x = 0$$

We can think of the expression on the left hand side as a polynomial in the variable $d/dt$ . We proceed by making the substitution $y = d/dt$ and then completing the square

$$y ^ {2} + \frac {b}{m} y + \frac {k}{m} = y ^ {2} + 2 \left(\frac {b}{2 m}\right) + \left(\frac {b}{2 m}\right) ^ {2} - \left(\frac {b}{2 m}\right) ^ {2} + \frac {k}{m} = \left(y + \frac {b}{2 m}\right) ^ {2} + \frac {k}{m} - \left(\frac {b}{2 m}\right) ^ {2}$$

So now our differential equation reads

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

where we have set

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

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

Now we just move one term to the other side to get

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

and we take the square root of this expression to get

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

Note that we now have two first order equations to solve (one for each sign).

We seek solutions to the equations

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

which have the obvious solutions

$$x = \exp \left(- \frac {b}{2 m} \pm i \omega\right) t = \exp \left(- \frac {b}{2 m} t\right) \exp (\pm i \omega t) = \exp \left(- \frac {b}{2 m} t\right) (\cos (\omega t) \pm i \sin (\omega t))$$

Thus our two solutions are (using Euler's formula)

$$x _ {1} = A _ {1} \exp \left(- \frac {b}{2 m} t\right) (\cos (\omega t) + i \sin (\omega t))$$ $$x _ {2} = A _ {2} \exp \left(- \frac {b}{2 m} t\right) (\cos (\omega t) - i \sin (\omega t))$$

and our total solution $(\mathbf{x}1 + \mathbf{x}2)$ can be written

$$x = \exp \left(- \frac {b}{2 m} t\right) \left(\left(A _ {1} + A _ {2}\right) \cos (\omega t) + i \left(A _ {1} - A _ {2}\right) \sin (\omega t)\right).$$

Now, we need to choose A1 and A2 so that we get a real-valued solution, that is $\mathsf{A}1 + \mathsf{A}2$ is real, and A1 - A2 is imaginary.

This condition has the effect of taking us from four unknown quantities (the real and imaginary part of each A) to just two, which is the appropriate number for a second order equation. Our solution is now

$$x = \exp \left(- \frac {b}{2 m} t\right) \big (B \cos (\omega t) + C \sin (\omega t) \big),$$

which is the general form of the solution representing damped oscillations, and we have

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

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

Then we rewrite our equation as

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

where we now have set

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

Then upon square rooting our equation we obtain

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

which is a real equation. The differential equation to solve is now

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

which has the solutions

$$x _ {1} = A _ {1} \exp \left(- \frac {b}{2 m} + \omega\right) t,$$ $$x _ {2} = A _ {2} \exp \left(- \frac {b}{2 m} - \omega\right) t,$$

both representing a damped motion without oscillations.