Simple Harmonic Oscillator

Problem

Figure 1. Mass on a spring harmonic oscillator

Consider the position function $x(t) = A \cos{(\omega t + \phi)}$.

Part 1: Determine the value of $\omega$ if this function solves the differential equation:

$$\frac{d^2 x}{dt^2}+\frac{kx}{m} = 0.$$

Part 2: Try to explain what each term of the above differential equation means.

Solution

Part 1

Take two time derivatives:

$$\frac{dx}{dt}= - \omega A \sin{(\omega t + \phi)}$$

$$\frac{d^2 x}{dt^2} = - {\omega}^{2} A \cos{(\omega t + \phi)}.$$

Consequently,

$$- {\omega}^{2} A \cos{(\omega t + \phi)} + \frac{k}{m} A \cos{(\omega t + \phi)} = 0$$

$$\left(\frac{k}{m}-{\omega}^{2}\right)A \cos{(\omega t + \phi)} = 0.$$

Divide away $A \cos{(\omega t + \phi)}$

$$\left(\frac{k}{m}-{\omega}^{2}\right) = 0$$

and solve to the angular frequency to get $\omega = \pm \sqrt{\frac{k}{m}}$.

This quantity is the angular frequency of the wave. Mathematically speaking, it is the eigenvalue of the differential equation.

Part 2

Multiply the entire equation by $m$:

$$m\frac{d^2 x}{dt^2} + kx = 0.$$

Since acceleration is the second time derivative of position $a = \frac{d^2 x}{dt^2}$

$$-kx = ma.$$

This is Newton’s second law with the spring force being the net force!