path: root/content/know/concept/harmonic-oscillator/index.pdc

---
title: "Harmonic oscillator"
firstLetter: "H"
publishDate: 2021-06-02
categories:
- Physics
- Mathematics

date: 2021-03-09T20:35:14+01:00
draft: false
markup: pandoc
---

# Harmonic oscillator

A **harmonic oscillator** obeys
the simple 1D version of [Hooke's law](/know/concept/hookes-law/):
to displace the system away from its equilibrium,
the needed force $F_d(x)$ scales linearly with the displacement $x(t)$:

$$\begin{aligned}
    F_d(x) = k x
\end{aligned}$$

Where $k$ is a system-specific proportionality constant,
called the **spring constant**,
since a spring is a good example of a harmonic oscillator,
at least for small displacements.
Hooke's law is also often stated for
the restoring force $F_r(x)$ instead:

$$\begin{aligned}
    F_r(x) = - k x
\end{aligned}$$

Let a mass $m$ be attached to the end of the spring.
After displacing it, we let it go $F_d = 0$,
so Newton's second law for the restoring force $F_r$ demands that:

$$\begin{aligned}
    F_r = m x''
\end{aligned}$$

But $F_r = - k x$,
meaning $m x'' = - k x$,
leading to the following equation for $x(t)$:

$$\begin{aligned}
    \boxed{
        x'' + \omega_0^2 x = 0
    }
\end{aligned}$$

Where $\omega_0 \equiv \sqrt{k / m}$ is the **natural frequency** of the system.
This differential equation has the following general solution:

$$\begin{aligned}
    \boxed{
        x(t)
        = C_1 \sin\!(\omega_0 t) + C_2 \cos\!(\omega_0 t)
    }
\end{aligned}$$

Where $C_1$ and $C_2$ are constants determined by the initial conditions.
For example, for $x(0) = 1$ and $x'(0) = 0$, the solution becomes:

$$\begin{aligned}
    x(t) = \cos\!(\omega_0 t)
\end{aligned}$$

When using Lagrangian mechanics or Hamiltonian mechanics,
we need to know the potential energy $V(x)$
added to the system by a displacement to $x$.
This equals the work done by the displacement,
and is therefore given by:

$$\begin{aligned}
    V(x) = \int_0^x F_d(x) \:dx = \frac{1}{2} k x^2 = \frac{1}{2} m \omega_0^2 x^2
\end{aligned}$$


## Damped oscillation

If there is a **friction force** $F_f$ affecting the system,
then the oscillation amplitude will decrease,
or it might not oscillate at all.
We define $F_f$ using a **viscous damping coefficient** $c$:

$$\begin{aligned}
    F_f = - c x'
\end{aligned}$$

Both $F_r$ and $F_f$ are acting on the system,
so Newton's second law states that:

$$\begin{aligned}
    m x'' = - c x' - k x
\end{aligned}$$

This can be rewritten in the following conventional form
by defining the **damping coefficient** $\zeta \equiv c / (2 \sqrt{m k})$,
which determines the expected behaviour of the system:

$$\begin{aligned}
    \boxed{
        x'' + 2 \zeta \omega_0 x' + \omega_0^2 x = 0
    }
\end{aligned}$$

The general solution is found from the roots $u$ of the auxiliary quadratic equation:

$$\begin{aligned}
    u^2 + 2 \zeta \omega_0 u + \omega_0^2 = 0
\end{aligned}$$

The discriminant $D = 4 \zeta^2 \omega_0^2 - 4 \omega_0^2$
tells us that the behaviour changes substantially
depending on the damping coefficient $\zeta$,
with three possibilities: $\zeta < 1$ or $\zeta = 1$ or $\zeta > 1$.

If $\zeta < 1$, there is **underdamping**:
the system oscillates with exponentially decaying
amplitude and reduced frequency $\omega_1 \equiv \omega_0 \sqrt{1 - \zeta^2}$.
The general solution is:

$$\begin{aligned}
    \boxed{
        x(t)
        = \big( C_1 \sin\!(\omega_1 t) + C_2 \cos\!(\omega_1 t) \big) \exp\!(- \zeta \omega_0 t)
    }
\end{aligned}$$

If $\zeta = 1$, there is **critical damping**:
the system returns to its equilibrium point in minimum time.
The general solution is given by:

$$\begin{aligned}
    \boxed{
        x(t)
        = \big( C_1 + C_2 t \big) \exp\!(- \omega_0 t)
    }
\end{aligned}$$

If $\zeta > 1$, there is **overdamping**:
the system returns to equilibrium slowly.
The general solution is as follows,
where $\omega_1 \equiv \omega_0 \sqrt{\zeta^2 - 1}$:

$$\begin{aligned}
    \boxed{
        x(t)
        = \big( C_1 \exp\!(\omega_1 t) + C_2 \exp\!(- \omega_1 t) \big) \exp\!(- \zeta \omega_0 t)
    }
\end{aligned}$$


## Forced oscillation

In the differential equations given above,
the right-hand side has always been zero,
meaning that the oscillator is not affected by any external forces.
What if we put a function there?

$$\begin{aligned}
    x'' + 2 \zeta \omega_0 x' + \omega_0^2 x = f(t)
\end{aligned}$$

Obviously, there exist infinitely many $f(t)$ to choose from,
and each needs a separate analysis.
However, there is one type of $f(t)$ that deserves special mention,
namely sinusoids:

$$\begin{aligned}
    \boxed{
        x'' + 2 \zeta \omega_0 x' + \omega_0^2 x = \frac{F}{m} \cos\!(\omega t + \chi)
    }
\end{aligned}$$

Where $F$ is a constant force, $\chi$ is an arbitrary phase,
and the frequency $\omega$ is not necessarily $\omega_0$.
We solve this case for $x(t)$ in detail.
Consider the complex version of the equation:

$$\begin{aligned}
    X'' + 2 \zeta \omega_0 X' + \omega_0^2 X = \frac{F}{m} \exp\!\big(i (\omega t + \chi)\big)
\end{aligned}$$

Then $x(t) = \Re\{X(t)\}$.
Inserting the ansatz $X(t) = C \exp\!(i \omega t)$,
for some constant $C$:

$$\begin{aligned}
    - C \omega^2 + C 2 i \zeta \omega_0 \omega + C \omega_0^2 = \frac{F}{m} \exp\!(i \chi)
\end{aligned}$$

Where $\exp\!(i \omega t)$ has already been divided out.
We isolate this equation for $C$:

$$\begin{aligned}
    C
    = \frac{F}{m \big((\omega_0^2 - \omega^2) + 2 i \zeta \omega_0 \omega\big)} \exp\!(i \chi)
    = \frac{F \big((\omega_0^2 - \omega^2) - 2 i \zeta \omega_0 \omega\big)}
    {m \big((\omega_0^2 - \omega^2)^2 + 4 \zeta^2 \omega_0^2 \omega^2\big)}
    \exp\!(i \chi)
\end{aligned}$$

We would like to rewrite this in polar form $C = r \exp\!(i \theta)$,
which turns out to be as follows:

$$\begin{aligned}
    C
    &= \frac{F}{m \sqrt{(\omega_0^2 - \omega^2)^2 + 4 \zeta^2 \omega_0^2 \omega^2}}
    \exp\!\bigg(i \chi - i \arctan\!\Big(\frac{2 \zeta \omega_0 \omega}{\omega_0^2 - \omega^2}\Big)\bigg)
\end{aligned}$$

For brevity, let us define the **impedance** $Z$
and the **phase shift** $\phi$
in the following way:

$$\begin{aligned}
    Z
    \equiv \sqrt{(\omega_0^2 - \omega^2)^2 / \omega^2 + 4 \zeta^2 \omega_0^2}
    \qquad \quad
    \phi
    \equiv \arctan\!\Big(\frac{2 \zeta \omega_0 \omega}{\omega_0^2 - \omega^2}\Big)
\end{aligned}$$

Returning to the original ansatz $X(t) = C \exp\!(i \omega t)$,
we take its real part to find $x(t)$:

$$\begin{aligned}
    \boxed{
        x(t)
        = \frac{F}{m \omega Z} \sin\!(\omega t + \chi - \phi)
    }
\end{aligned}$$

Two things are noteworthy here.
Firstly, $f(t)$ and $x(t)$ are out of phase by $\phi$; there is some lag.
This is caused by damping, because if $\zeta = 0$, it disappears $\phi = 0$.

Secondly, the amplitude of $x(t)$ depends on $\omega$ and $\omega_0$.
This brings us to **resonance**,
where the amplitude can become extremely large.
Actually, resonance has two subtly different definitions,
depending on which one of $\omega$ and $\omega_0$ is a free parameter,
and which one is fixed.

If the natural $\omega_0$ is fixed and the driving $\omega$ is variable,
we find for which $\omega$ resonance occurs by minimizing the amplitude denominator $\omega Z$.
We thus find:

$$\begin{aligned}
    0
    = \dv{(\omega Z)}{\omega}
    = \frac{- 4 \omega_0^2 \omega + 4 \omega^3 + 8 \zeta^2 \omega_0^2 \omega}{2 \sqrt{(\omega_0^2 - \omega^2)^2 + 4 \zeta^2 \omega_0^2 \omega^2}}
    \quad \implies \quad
    \boxed{
        \omega = \omega_0 \sqrt{1 - 2 \zeta^2}
    }
\end{aligned}$$

Meaning the resonant $\omega$ is lower than $\omega_0$,
and resonance can only occur if $\zeta < 1 / \sqrt{2}$.

However, if the driving $\omega$ is fixed and the natural is $\omega_0$ is variable,
the problem is bit more subtle:
the damping coefficient $\zeta = c / (2 m \omega_0)$
depends on $\omega_0$.
This leads us to:

$$\begin{aligned}
    0
    = \dv{(\omega Z)}{\omega_0}
    = \frac{4 \omega_0^3 - 4 \omega^2 \omega_0}{2 \sqrt{(\omega_0^2 - \omega^2)^2 + c^2 \omega^2 / m^2}}
    \quad \implies \quad
    \boxed{
        \omega_0 = \omega
    }
\end{aligned}$$

Surprisingly, the damping does not affect $\omega_0$, if $\omega$ is given.
However, in both cases, the damping *does* matter for the eventual amplitude:
$c \to 0$ leads to $x \to \infty$,
and resonance disappears or becomes negligible for $c \to \infty$.



## References
1.  M.L. Boas,
    *Mathematical methods in the physical sciences*, 2nd edition,
    Wiley.