path: root/source/know/concept/harmonic-oscillator/index.md

---
title: "Harmonic oscillator"
sort_title: "Harmonic oscillator"
date: 2021-06-02
categories:
- Physics
- Mathematics
layout: "concept"
---

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](/know/concept/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:

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

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 $$x(t) = \Real\{X(t)\}$$.
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.