The Harmonic Oscillator: A classical overview

Hooke first discovered the law of springs, that their force is proportional to their displacement. In reality, their force is generally a far more complicated function. In fact, many complicated forces can be approximated in a form similar to the harmonic oscillator. Consider the potential $U(x)$ describing some physical phenomenon. Close enough to the equilibrium point x=a, we can expand this function as a Taylor polynomial,

$$U(x)=U(a) + \overbrace{\left(\frac{\partial U(x)}{\partial x} = 0... ...rbrace{\left(\frac{\partial ^2 U(x)}{\partial x^2}\right)}^{x=a}(x-a)^2+ \cdots$$ (1.1)

Here the first derivative is zero, as we take the Taylor series around an equilibrium. How close is close enough? This would depend on the degree of accuracy one seeks, though generally it is assumed that

$$\frac{\left(\frac{\partial^3U(x)}{\partial x^3}\right)(x-a)^3}{\left(\frac{\partial^2U(x)}{\partial x^2}\right)(x-a)^2} \ll 1$$

We are at liberty to rescale our potential so that $U(a)=0$, whereby we have the approximation,

$$U(x)=\frac{1}{2!}\left(\frac{\partial^2U(x)}{\partial x^2}\right)(x-a)^2$$ (1.2)

It is typical to also rescale the x-axis so that $a=0$ and, using common terminology,

$$U(x)=\frac{1}{2!}\left(\frac{\partial^2U(x)}{\partial x^2}\right)x^2 \equiv \frac{1}{2}kx^2$$

Here k is a constant. The equation of motion for such an object becomes

$$\frac{dp}{dt}=-kx, \$$   typically$$\ \ m\frac{d^2x}{dt^2}+kx=0$$

Often it is more accurate to add a damping term, for example, a spring under water is better described by,

$$m\frac{d^2x}{dt^2}+\nu \frac{d x}{dt} + kx = 0$$ (1.3)

Here $\nu$ is some constant defined by the retarding force $F_r = -\nu \frac{dx}{dt}$. If a harmonic driving force is applied, Equation 1.3 becomes inhomogeneous,

$$m\frac{d^2x}{dt^2}+\nu \frac{d x}{dt} + kx = F_0\cos(\omega t)$$ (1.4)

Equation 1.3 is easiest to solve. Let $\nu/m = \gamma$, $F_0/m \rightarrow A$, and $\omega_0^2=k/m$. If we suppose the solution $x=\exp(rt)$,

$$x'' + \gamma x' + \omega_0^2x = 0 \ \rightarrow \ r^2 +\gamma r + \omega_0^2 = 0 \ \ni r=\frac{-\gamma \pm \sqrt{\gamma^2 -4\omega_0^2}}{2}$$ (1.5)

We seek a fundamental set of solutions to these equations. From the theory of differential equations we know that we require exactly two distinct solutions. The $\pm$ in the solution to the characteristic equation lends us two such solutions, except in the case when $\gamma^2 = 4\omega_0^2$. In that case, our first solution is

$$x_1(t)=e^{-\gamma t/2}$$ (1.6)

The characteristic equation suggest no further solution. The method of reduction of order has us assume that

$$x(t)=v(t)x_1(t) = v(t)e^{-\gamma t/2}$$ (1.7)

Plugging Equation 1.7 into Equation 1.5, we find that $v''(t)=0$ and so

$$v(t)=c_1 t + c_2 \ \ni \ x(t)=c_1 t e^{-\gamma t/2} + c_2 e^{-\gamma t/2}$$

The Wronskian of these two solutions is nonzero always, and so we have a fundamental set of solutions. The other cases are more easily solved and we find,

$$x(t)= \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad$$

$$\left\{\begin{array}{ll} a_1 e^{-\gamma t/2}\exp(t\sqrt{\gamma^2 ... ...t\sqrt{-\gamma^2 +4 \omega_0^2}/2) & \gamma -4\omega_0^2 < 0 \end{array}\right.$$

The latter can be written,

$$x(t)=d_1e^{-\gamma t/2}\cos(t\sqrt{-\gamma^2 +4 \omega_0^2}/2)+d_2e^{-\gamma t/2}\sin (t\sqrt{-\gamma^2 +4 \omega_0^2}/2)$$

The driven Equation (1.4) is slightly more complicated to solve. In Equation 1.4 one can replace $\cos(\omega t)$ with $\exp(i\omega t)$ on the condition we take the real part of the solution to be found from this ``axillary equation'' [2]. If we suppose a solution $x_d(t)=A\exp(i(\omega t + \phi))$, we find,

$$\left(-\omega^2 A + \gamma i \omega A + \omega_0^2 A\right)e^{i(\omega t + \phi)} = (F_0/m)e^{i\omega t}$$ (1.8)

Dividing out the exponent on the left hand side and equating real and imaginary parts of the right yields,

$$\frac{A(w_0^2 - \omega^2)}{F_0/m}=\cos \phi, \ \frac{\gamma \omeg... ...i=\tan^{-1}\left(\frac{\gamma \omega}{\omega^2 - \omega_0^2}\right) \end{array}$$

The full solution to the driven damped harmonic oscillator is then,

$$X(t)=x(t)+x_d(t)$$ (1.9)

Here x(t) is the solution to the homogeneous equation we found above (one of the three variations depending on the parameters). Or, since we enjoy seeing large equations in their full forms,

$$X(t)= \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad$$

$$\left.\begin{array}{l}a_1 e^{-\gamma t/2}e^{t\sqrt{\gamma^2 -4 \o... ...^2 - \omega_0^2}\right) )}}{m\sqrt{(\omega_0^2 - \omega^2)^2+\gamma^2\omega^2}}$$

Figure 1.1: Examples of three possible solution types for the homogeneous damped harmonic equation.