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.]


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