Harmonic Oscillator

#physics
A quick recap of the harmonic oscillator we are familiar with through JEE.

We have a mass, connected to a spring. The equation of motion describing this is $m \ddot{x} = - k x$
The solution to this is $$x = B\sin(\omega_{0}t) + C\cos(\omega_{0}t)$$
where $$\omega_{0} = \sqrt{ \frac{k}{m} }$$
In general, $x$ is written as $$x = A\cos(\omega t + \phi)$$where $$\phi = \tan^{-1}\left( -\frac{B}{C} \right)$$
Total energy of this harmonic oscillator is constant, $$E = \frac{1}{2}kA^2$$where $A$ is the amplitude of the oscillation, measured from mean position to extreme position.

We denote the time average value of any function as $$\langle f\rangle$$ and therefore $$\begin{align} \langle U \rangle = \frac{1}{4}kA^2 \ \ \langle K\rangle = \frac{1}{4}kA^2 \end{align}$$
where $U$ and $K$ denote potential and kinetic energy of the oscillator, respectively.

Damped harmonic oscillator

We consider our discussion with a viscous force, modeled of the type $$f = -bv$$where $v$ is the velocity of the mass, and $b$ is a proportionality factor that depends on the shape of the mass and medium in which it is oscillating.

The equation of motion can be written as $$\ddot{x} + 2\gamma \dot{x} + \omega_{0}^2x = 0 $$
here, $ω_{0}^{2} = k / m$ and $γ = b / 2 m$ .
The knowledge of how to solve this is given in Homogeneous Second Order ODE
We obtain the final solution as $$x(t) = e^{-\gamma t}[Ae^{\Omega t} + Be^{-\Omega t}]$$where $$\Omega^2 = \gamma^2 - \omega_{0}^2$$
This leads to three cases

$Ω^{2} < 0$ : Underdamping. The most general case
$Ω^{2} = 0$ : Critically damped
$Ω^{2} > 0$ : Overdamped

Underdamped

We find that $Ω = i \sqrt{ω_{0}^{2} - γ^{2}}$ and from our pre-existing knowledge of differential equations, we can say that the general solution for the linear differential equation of motion are simply : $$x(t) = e^{-\gamma t}[C_{1}\cos(\Omega t) + C_{2}\sin(\Omega t)]$$which is equivalent to combining the terms inside the bracket and saying the solution is $$x(t) = Ae^{-\gamma t}\cos(\Omega t + \phi)$$
Therefore, the oscillation decays as $$A_{0}e^{-\gamma t}$$where $A_{0}$ is the initial amplitude.

For the following part, we proceed with $ω_{1} = Ω$ , and $γ = b / m$

If we plot the motion of the graph, the amplitude falls of exponentially as dictated by $A (t)$ but the points where it crosses the $x$ axis are spaced equally, separated by $2 π / ω_{1}$ . It is for this reason that $ω_{1}$ is frequently called the "frequency of oscillation".

Qualitative analysis

The essential features of the motion depend on the quantity $γ / ω_{1}$

If it is $≪ 1$ , then $A (t)$ decreases very little and the damping is not very prominent. This is called the "lightly damped" condition.
If it is rather large, $A (t)$ tends to decrease quite rapidly, and this case is called "Heavily damped"
For light damping $ω_{1} \approx ω_{0}$ , but for heavy damping, $ω_{0}$ can be quite larger than $ω_{1}$

Energy consideration

The work done by the resistive viscous force is non-zero, and therefore it is obvious that the energy of the system must decrease with time.

W_{fric} = \int_{0}^{t} f v d t = - \int_{0}^{t} b v^{2} d t

Therefore energy is dissipated away.

We attempt to find the expressions for energy in the limit that $γ / ω_{1} ≪ 1$
Skipping all the derivation (Read Kleppner Kolenkow for details)
we finally obtain $$E(t) = \frac{1}{2}kA^2\exp(-\gamma t)$$
A remarkably simple result, it shows that energy falls of approximately exponentially in a lightly damped system.

The time constant $τ$ for this decay is $$\tau = \frac{1}{\gamma} = \frac{m}{b}$$
$τ$ is often called the damping time, or alternatively, characteristic time of the system

Q value of an oscillator

The degree of damping of an oscillator is often represented by a dimensionless parameter $Q$ called the Quality factor. $$Q = \frac{\pu{Energy stored in the oscillator}}{\pu{Energy dissipated per radian}} = 2\pi\frac{\pu{Energy stored in the oscillator}}{\pu{Energy dissipated per cycle}}$$
The time to oscillate through one radian is $1 / ω_{1}$

For the lightly damped case, $Q$ is easily calculated. The rate of change of energy is $$ -\gamma E$$
and combining it with the time, $$Q = \frac{E}{\gamma E/\omega_{1}} = \frac{\omega_{1}}{\gamma} \approx \frac{\omega_{0}}{\gamma}$$
A lightly damped oscillator has $Q ≫ 1$ , a heavily damped oscillator has low $Q$ .

Critically damped

again, refer Homogeneous Second Order ODE
The general solution in this case is $$x(t) = e^{-\gamma t}(C + Dt)$$

Overdamping

The general solution in this case is $$x(t) = e^{-\gamma t}[Ae^{\Omega t} + Be^{-\Omega t}]$$

Note

Critically damped is generally referred to as the fastest way to return to equilibrium
Neither of the critically damped or overdamped cases ever really reaches 0 amplitude, takes infinite time.
However, critically damped manages to minimize it's value faster, and reaches any error $ϵ > 0$ around equilibrium faster than overdamped.

Energy and power

For the damped harmonic oscillator, the resistive force does work over time and decays the energy of the mass.
It can be given by the following $$\begin{gather} E_{tot} = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}kx^2 \ \ \frac{dE_{tot}}{dt} = m\dot{x}\ddot{x} + kx\dot{x} = m\dot{x}(\ddot{x} + \omega_{0}^2x) \ \ \boxed{\frac{dE_{\pu{ tot }}}{dt} = - m\gamma \dot{x}^2} = \pu{ Power } \end{gather}$$

Forced Harmonic Oscillator