Oscillations and Simple Harmonic Motion

The Qualitative Physics Of Oscillation

In the slow motion movie clip at right, the mass glides on an air track. The track is perforated with small holes, through which flows air from the inside, where the pressure is above atmospheric. So the mass is supported, like a hovercraft, on a cushion of air, and friction is eliminated. Because the speed is small, air resistance is very small. Consequently, the only non-negigible force in the horizontal direction is that exerted by the two springs. Because there are no vertical displacements, we discuss here only the horizontal displacement.

At the equilibrium position (x = 0 in the graph below the clip), the forces exerted by the two springs are equal in magnitude but opposite in direction, so the total force is zero. To the right of equilibrium, the force acts to accelerate the mass to the left, and vice versa. (The graph is rotated 90° from its normal orientation so that we can compare it with the motion.)

Let's begin (as do the graph and the animation) with the mass to the right of equilibrium and at rest. Let's see what happens when I release it:

First, the spring force acts to the left and mass is accelerated towards x = 0.
When it reaches x = 0, it has a velocity and therefore a momentum to the left. (Near equilbrium, the forces are small, so there is a region near x = 0 over which the velocity changes little: the x(t) graph is almost straight.)
When it arrives at x = 0, because of its momentum to the left, it overshoots, i.e. it continues travelling to the left. While it is to the left of x = 0, however, the spring force acts to the right. This force gradually slows the mass until it stops. The point at which it stops is, of course, its maximum displacement to the left.
Once it is stopped on the left hand side of equilibrium, the spring force accelerates it to the right, so the velocity and momentum to the right increase.
When it reaches equilibrium again, it now has its maximum rightwards momentum.
It overshoots and continues to the right. The spring force now acts to the left, so it decelerates until it stops at its maximum rightwards displacement.
Because no non-negligible nonconservative forces act, mechanical energy is conserved. Consequently, the system returns to its initial condition. The cycle then repeats exactly, so the motion is periodic.

Quantitative Analysis
For linear springs, this leads to Simple Harmonic Motion. The force F exerted by the two springs is F = − kx, where k is the combined spring constant for the two springs (see Young's modulus, Hooke's law and material properties). In this case, k = k1 + k2, where k1 and k2 are the constants of the two springs. The analysis that follows here is fairly brief. However, we do a quantitative analysis on the multimedia chapter Oscillations and also solve this problem as an example on Differential Equations. There is also a page on the Kinematics of Simple Harmonic Motion.
Newton's second law states that the acceleration d2x/dt2 of the mass m subject to total force F satisfies F = m.d2x/dt2 , which gives the differential equation

m.d2x/dt2 = − kx, or
d2x/dt2 = − ω2 x , where ω2 = k/m .
Solving this particular equation is described in detail on the Differential Equations page. However, we can verify by subsitution that the solution is
x = A sin (ωt + φ),

where A is the amplitude, and the phase constant φ is determined by the initial conditions. We discuss these below.

Initial Conditions
In the first movie shown shown at right, the mass is released from rest, so the amplitude is maximal (x = A) at t = 0, so the required phase constant is φ = π/2. (Indeed, for this particular case, we could say that the curve is a cos function rather than a sine.)

x1 = A sin (ωt + π/2) = A cos (ωt )



In the second movie shown at right, however, the mass is given an impulsive start, so the initial condition approximates maximum velocity and x = 0 at t = 0. This requires φ = 0, so

x2 = A sin (ωt + 0) = A sin (ωt)

Here we start with an initial velocity, which is

v0 = dx2/dt = A sin ωt = Aω
Note that the initial condition determines both φ and A.
In both these clips, a rotating line (an animated phasor diagram) is used to show that Simple Harmonic Motion is the projection onto one dimension of circular motion. This is explained in detail in the Kinematics of Simple Harmonic Motion in Physclips.

Phasors are commonly used to facilitate calculations in AC circuits.

Frequency f And Angular Frequency ω
We saw above that x = A sin (ωt + φ), where ω2 = k/m . The cyclic frequency is f = 1/T, where T is the period. The sine function goes through one complete cycle when its argument increases by 2π, so we require that (ω(t+T) + φ) − (ωt + φ) = 2π, so ωT = 2π, so

ω = 2π/T = 2πf = (k/m)½ .
This parameter is determined by the system: the particular mass and spring used. For a linear system, the frequency is independent of amplitude (see below, however, a for nonlinear system ).

Compare the oscillations shown in the two clips at right. The first uses one air track glider and the second uses two similar gliders, so the mass is doubled. The period is increased by about 40%, i.e. by a factor of √2, so the frequency is decreased by the same factor.




Though it is not so easy to see in the video, at right we have used stiffer springs with a higher value of k. Here the period is shorter and therefore the frequency higher that in all the previous examples.