PHY5200 F06

Chapter 5: Oscillations

Reading

Taylor 5.1-5.2 (today) and midterm on Friday.

Recall

Any potential energy function can be expanded in a Taylor series about a point of stable equilibrium. Being a point of stable equilibrium, the first order term of the expansion is zero, so the first term that contributes to the force is the second order term. Neglecting higher order terms, the resulting force is the spring-like linear restoring force.

Example: The Simple Pendulum

A simple pendulum consists of a mass m suspended from a massless string of length l. The string is attached to a support (ceiling) and the system is free to oscillate (in a plane for now). Show that for small oscillations, the mass is subject to a linear restoring force.

The potential energy is all gravitational potential energy, given by

U = mgh = mgl(1-cosθ)

where θ=0 when the string is vertical. For small oscillations, expand U around θ=0 (up to second order in θ)

U ≅ ½mgl&theta² + ... ≅ ½m(g/l)x²

where x = lθ is the displacement from equilibrium. The corresponding force is

F = -∂U/∂x = -m(g/l)x

which is a linear restoring force with spring constant k = mg/l.

Simple Harmonic Motion

Let's now investigate the motion of an object subject to a linear restoring force. When the linear restoring force is the only (relevant) force in the problem, the result is known as simple harmonic oscillation. Another way to state this is that the potential energy is quadratic in the displacement, to within a constant, U = U0 + ½kx². These are equivalent since F = -∂U/∂x = -kx. There are a number of ways to express the result, and we'll explore four common ways here.

The equation of motion for the simple harmonic oscillator is

mx[ddot] = -kx    or    x[ddot] = -(k/m)x = -ω²x

This is a linear, homogeneous, second order differential equation. We expect two independent constants in the solution. The equations says that the position is described by a function whose second derivative is proportional to the original function. Functions with this property are sine, cosine, and exponential.

The Exponential Solutions

Let's begin by trying the exponential function. Make a trial solution of X(t) = Ceαt. By the above, we don't mean that the position is an imaginary number, but rather we will take the real (or imaginary) part of the resulting function as the result. Taking the second derivative and inserting into the differential equation yields:

α²Ceαt = -ω²Ceαt

Notice that the exponential factors containing the time dependence are the same, and since the exponential is never always zero, so we can factor it out and get the relation

α² = -ω²

and, since k and m are positive real numbers we know that ω² is a positive real number, therefore α = ±iω. This is two solutions, +ω and -ω, each with an arbitrary constant. Putting these pieces together we can write the solution in exponential form:

X(t) = C1eiωt + C2e-iωt

Again, x is not a complex number. One must take the real or imaginary part of this expression for the value of x, that is:

x(t) = Re[X(t)]   or   x(t) = Im[X(t)].

Note that C1 and C2 can be complex numbers.

The Sine and Cosine Solutions

One could have started with a sine or cosine trial solution as well. The exponential solution can be transformed into the sine/cosine solution by writing

X(t) = ½B1(eiωt + e-iωt) - i½B2(eiωt - e-iωt)
x(t) = B1cos(ωt) + B2sin(ωt)

The Phase-Shifted Cosine Solution

x(t) = A cos(ωt - δ)

Solution as the Real Part of a complex Exponential

x(t) = Re( Ceiωt ) = Re( Aei(ωt - δ) )

© Robert Harr 2006