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 Fd(x)F_d(x) scales linearly with the displacement x(t)x(t):

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

Where kk 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 Fr(x)F_r(x) instead:

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

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

Fr=mx\begin{aligned} F_r = m x'' \end{aligned}

But Fr=kxF_r = - k x, meaning mx=kxm x'' = - k x, leading to the following equation for x(t)x(t):

x+ω02x=0\begin{aligned} \boxed{ x'' + \omega_0^2 x = 0 } \end{aligned}

Where ω0k/m\omega_0 \equiv \sqrt{k / m} is the natural frequency of the system. This differential equation has the following general solution:

x(t)=C1sin(ω0t)+C2cos(ω0t)\begin{aligned} \boxed{ x(t) = C_1 \sin(\omega_0 t) + C_2 \cos(\omega_0 t) } \end{aligned}

Where C1C_1 and C2C_2 are constants determined by the initial conditions. For example, for x(0)=1x(0) = 1 and x(0)=0x'(0) = 0, the solution becomes:

x(t)=cos(ω0t)\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)V(x) added to the system by a displacement to xx. This equals the work done by the displacement, and is therefore given by:

V(x)=0xFd(x)dx=12kx2=12mω02x2\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 FfF_f affecting the system, then the oscillation amplitude will decrease, or it might not oscillate at all. We define FfF_f using a viscous damping coefficient cc:

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

Both FrF_r and FfF_f are acting on the system, so Newton’s second law states that:

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

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

x+2ζω0x+ω02x=0\begin{aligned} \boxed{ x'' + 2 \zeta \omega_0 x' + \omega_0^2 x = 0 } \end{aligned}

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

u2+2ζω0u+ω02=0\begin{aligned} u^2 + 2 \zeta \omega_0 u + \omega_0^2 = 0 \end{aligned}

The discriminant D=4ζ2ω024ω02D = 4 \zeta^2 \omega_0^2 - 4 \omega_0^2 tells us that the behaviour changes substantially depending on the damping coefficient ζ\zeta, with three possibilities: ζ<1\zeta < 1 or ζ=1\zeta = 1 or ζ>1\zeta > 1.

If ζ<1\zeta < 1, there is underdamping: the system oscillates with exponentially decaying amplitude and reduced frequency ω1ω01ζ2\omega_1 \equiv \omega_0 \sqrt{1 - \zeta^2}. The general solution is:

x(t)=(C1sin(ω1t)+C2cos(ω1t))exp(ζω0t)\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 ζ=1\zeta = 1, there is critical damping: the system returns to its equilibrium point in minimum time. The general solution is given by:

x(t)=(C1+C2t)exp(ζω0t)\begin{aligned} \boxed{ x(t) = \big( C_1 + C_2 t \big) \exp(- \zeta \omega_0 t) } \end{aligned}

If ζ>1\zeta > 1, there is overdamping: the system returns to equilibrium slowly. The general solution is as follows, where ω1ω0ζ21\omega_1 \equiv \omega_0 \sqrt{\zeta^2 - 1}:

x(t)=(C1exp(ω1t)+C2exp(ω1t))exp(ζω0t)\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?

x+2ζω0x+ω02x=f(t)\begin{aligned} x'' + 2 \zeta \omega_0 x' + \omega_0^2 x = f(t) \end{aligned}

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

x+2ζω0x+ω02x=Fmcos(ωt+χ)\begin{aligned} \boxed{ x'' + 2 \zeta \omega_0 x' + \omega_0^2 x = \frac{F}{m} \cos(\omega t + \chi) } \end{aligned}

Where FF is a constant force, χ\chi is an arbitrary phase, and the frequency ω\omega is not necessarily ω0\omega_0. We solve this case for x(t)x(t) in detail. Consider the complex version of the equation:

X+2ζω0X+ω02X=Fmexp ⁣(i(ωt+χ))\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)}x(t) = \Real\{X(t)\}. Inserting the ansatz X(t)=Cexp(iωt)X(t) = C \exp(i \omega t), for some constant CC:

Cω2+C2iζω0ω+Cω02=Fmexp(iχ)\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(iωt)\exp(i \omega t) has already been divided out. We isolate this equation for CC:

C=Fm((ω02ω2)+2iζω0ω)exp(iχ)=F((ω02ω2)2iζω0ω)m((ω02ω2)2+4ζ2ω02ω2)exp(iχ)\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=rexp(iθ)C = r \exp(i \theta), which turns out to be as follows:

C=Fm(ω02ω2)2+4ζ2ω02ω2exp ⁣(iχiarctan ⁣(2ζω0ωω02ω2))\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 ZZ and the phase shift ϕ\phi in the following way:

Z(ω02ω2)2/ω2+4ζ2ω02ϕarctan ⁣(2ζω0ωω02ω2)\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)=Cexp(iωt)X(t) = C \exp(i \omega t), we take its real part to find x(t)x(t):

x(t)=FmωZsin(ωt+χϕ)\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)f(t) and x(t)x(t) are out of phase by ϕ\phi; there is some lag. This is caused by damping, because if ζ=0\zeta = 0, it disappears ϕ=0\phi = 0.

Secondly, the amplitude of x(t)x(t) depends on ω\omega and ω0\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 ω\omega and ω0\omega_0 is a free parameter, and which one is fixed.

If the natural ω0\omega_0 is fixed and the driving ω\omega is variable, we find for which ω\omega resonance occurs by minimizing the amplitude denominator ωZ\omega Z. We thus find:

0=d(ωZ)dω=4ω02ω+4ω3+8ζ2ω02ω2(ω02ω2)2+4ζ2ω02ω2    ω=ω012ζ2\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 ω\omega is lower than ω0\omega_0, and resonance can only occur if ζ<1/2\zeta < 1 / \sqrt{2}.

However, if the driving ω\omega is fixed and the natural is ω0\omega_0 is variable, the problem is bit more subtle: the damping coefficient ζ=c/(2mω0)\zeta = c / (2 m \omega_0) depends on ω0\omega_0. This leads us to:

0=d(ωZ)dω0=4ω034ω2ω02(ω02ω2)2+c2ω2/m2    ω0=ω\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 ω0\omega_0, if ω\omega is given. However, in both cases, the damping does matter for the eventual amplitude: c0c \to 0 leads to xx \to \infty, and resonance disappears or becomes negligible for cc \to \infty.


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