# Classical Simple Harmonic Oscillators.

## Mass on a Spring System. Problem statement

Let m be the mass of a structureless body supported by a spring
with a uniform force constant k as shown in the diagram.
Set up the differential equation of motion that determines
the displacement of the mass from its equilibrium position
at time t when the intital conditions are x(0) = x0 and x'(0) = 0.

## Solution of the problem

The net force on the mass is given by Hooke's law: f = -kx, where x is the extension of the spring beyond its equilibrium length. According to Newton's second law the force can be equated to mass times acceration giving us the second order DE,
mx" + kx = 0.

To solve, define w = (k/m)1/2 and rewrite the DE as
x" + w2x = 0.

The complete solution of the DE, with two arbitrary constants, is
x(t) = A cos(w t) + B sin(w t).
The initial conditions (x(0)=x0 and x'(0)=0) require that B = 0, and A = x0.

## Discussion

The constant w, characteristic of the system mass and force constant, is the angular frequency of the motion. The frequency in cycles per unit time is n = w/2p, and the period is T = 2p/w. Motion of this type, with amplitude given by the cosine (or the sine) is called harmonic. When the initial conditions are different, the solution can still be expressed as a cosine using the alternative parameterization x(t) = x0 cos(w t - f). This choice of parameters includes the phase angle f.

## Electrical Analog. LC oscillator

A physically different system with an equivalent DE and analogous "motion" consists of a series electric circuit contaning capacitance C and inductance L as shown in the figure. In this, the charge on the capacitor, C, varies in time in the same way as the displacement of the mass on a spring.

## Problem Setup

Suppose that, with the switch in the open position, the capacitor has charge Q0 and the current, I, is zero. Then, at time t=0, the switch is closed allowing charge to flow. Let the circuit have no resistance and find the capacitor charge Q(t) at later times.

## Solution

The potential differences across circuit elements must sum to zero around the closed circuit. That across the inductance is L dI/dt and that across the capacitor is Q/C. Since I = dQ/dt we find the DE: LQ" + Q/C = 0. Make the substitution w2 = 1/LC so that the DE becomes Q" + w2 Q = 0. The complete solution is Q(t) = A cos(w t) + B sin(w t) or in a different parameterization Q(t) = C cos(wt - f). The initial conditions determine the arbitrary constants giving Q(t) = Q0 cos(w t).

This result is identical in form with that for the mass on a spring. Therefore the electrical system and the mechanical system are said to be analogous. The equivalence between them is exhibited in x ~ Q, m ~ L, k ~ 1/C. That is, displacement is equivalent to quantity of electrical charge, mass to inductance, and spring constant to the reciprocal of capacitance.

## Exercises

1. Show that the different parameterizations x(t) = A cos(w t) + B sin(w t) and x(t) = C cos(w t - f) are related by C = (A2 + B2)1/2 and f = tan-1(B/A).

2. Deduce the form of the DE for the electrical LC oscillator if resistance, R, is added to the series circuit. Only the ratios R/L=2k and 1/LC=w2 are significant for the DE. (a) To solve, introduce the integrating factor, Q(t) = u(t) exp(-kt). (b) Show that there are three cases with distinctly different DE: (i) w > k , (ii) w < k, and (iii) w = k. (c) Solve for Q(t) in this in each case. [Case (i) with Q(t) = e-kt C cos(at - f) is called the damped oscillator. Here a2 = w2 - k2, and C and f are arbitrary constants determined by initial conditions.]

Home
Created or up-dated 08/03/99   by R.D. Poshusta