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.

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