The Simple Harmonic Oscillator

Simple Harmonic Motion For SHM to occur we require stable equilibrium, about a point. For example, at the origin we could have: ∑ F ⃗ ( 0 ) = 0 , which would describe a system in equilibrium. This however is not necessarily stable equilibrium.

A simple cartoon of stable and unstable equilibrium. The lower part of the figure shows the case of unstable equilibrium. The upper part shows the case of stable equilibrium. These situations often occur in mechanical systems. The lower part of the figure shows the case of unstable equilibrium. The upper part shows the case of stable equilibrium. These situations often occur in mechanical systems. For example, consider a mass attached to a spring:

In general, in a case of stable equilibrium we can write the force as a polynomial expansion: F ( x ) = − ( k 1 x + k 2 x 2 + k 3 x 3 + … ) where the k i are positive constants. There is always a region of x small enough that we can write: F = − k x F = − k x m a = − k x m x ¨ = − k x x ¨ + k m x = 0 This is satisfied by an equation of the form x = A sin ( ω t + φ 0 ) where A and φ 0 are constants that are determined by the initial conditions. Draw a diagram of a sinusoid and mark on it the period T and Amplitude A

φ 0 Is an arbitrary phase which shifts the sinusoid.This is also satisfied by an equation of the form x = A sin ( ω t ) + B cos ( ω t ) Lets show this: x = A sin ( ω t ) + B cos ( ω t ) x ˙ = ω ( A cos ( ω t ) − B sin ( ω t ) ) x ¨ = − ω 2 ( A sin ( ω t ) + B cos ( ω t ) ) x ¨ = − ω 2 x Again there are two constants determined by the initial conditions A and B The equation can be rewritten x ¨ + ω 2 x = 0 Thus if ω 2 = k m then the equation is identical to the SHM equation. So another way to write the equation of Simple Harmonic Motion is x ¨ + ω 2 x = 0 or x ¨ = − ω 2 x It is also important to remember the relationships between freqency, angular frequency and period: ω = 2 π ν T = 2 π ω ν = 1 T Another solution to the SHM equation is x ˜ = A cos ( ω t + φ 0 ) + i A sin ( ω t + φ 0 ) Recall Taylor's expansions of sine and cosine sin θ = θ − θ 3 3 ! + θ 5 5 ! … cos θ = 1 − θ 2 2 ! + θ 4 4 ! … Then cos θ + i sin θ = 1 + i θ − θ 2 2 ! − i θ 3 3 ! + θ 4 4 ! … = 1 + i θ + ( i θ ) 2 2 ! + ( i θ ) 3 3 ! + ( i θ ) 4 4 ! … = e i θ (an alternative way to show this is the following) z ≡ cos θ + i sin θ ⅆ z = ( − sin θ + i cos θ ) ⅆ θ = i z ⅆ θ ∫ ⅆ z z = ∫ i ⅆ θ ln z = i θ z = e i θ Thus we can write x ˜ = A cos ( ω t + φ 0 ) + i A sin ( ω t + φ 0 ) as x ˜ = A e i ( ω t + φ 0 ) x ˜ = A e i ( ω t + φ 0 ) x ˜ ˙ = i ω A e i ( ω t + φ 0 ) x ˜ ¨ = ( i ω ) 2 A e i ( ω t + φ 0 ) = − ω 2 x ˜ We will use the complex representation a lot, so you need to become familiar with it. It is used a lot in Optics, Classical and Quantum Mechanics and Electrical Engineering so it is a good thing to know. Now for physical systems we are interested in just the real part so x = R e [ A e i ( ω t + φ 0 ) ] This will be implicitly understood. In physics we just write x = A e i ( ω t + φ 0 ) One thing that will seem to be confusing is that there are all these different solutions. They are all just different forms of the same thing. Which form is used in a particular circumstance is simply a matter of convenience. Some forms lend themselves to to solutions of certain problems more easily than others. Also the most convenient form can depend upon the initial conditions. For example if x is at its maximum displacement at time t = 0 then a cos form may be the most convenient. As a general rule I like using the complex representation because natural logarithms are so easy to work with. For example ⅆ e x ⅆ x = e x ⅆ e a x ⅆ x = a e a x ∫ e a x ⅆ x = 1 a e a x which is all pretty simple to remember