Categories: Mathematics, Physics.

# Harmonic oscillator

A harmonic oscillator obeys the simple 1D version of Hooke’s 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 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 $$c \to \infty$$.

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