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WannabeNewton
#10
Feb12-13, 11:42 AM
C. Spirit
Sci Advisor
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The general solution to the DE [itex]r'' + \omega ^{2}r = 0[/itex] would be [itex]r(t) = Ae^{i\omega t} + Be^{-i\omega t} [/itex]. For simplicity of visualization, take A = 1, B = 0 so that [itex]r(t) = e^{i\omega t}[/itex]. You will recognize this as being [itex]exp:I\rightarrow S^{1}[/itex] where [itex]S^{1}[/itex] is the unit circle (we are viewing it as being a subset of the complex plane) and [itex]I[/itex] is an appropriate interval in [itex]\mathbb{R}[/itex]. Now, with regards to the physics of simple harmonic motion, we just take the real part of the above expression for [itex]r(t)[/itex] but you can visualize [itex]\omega [/itex], the angular frequency, as being the rate at which the unit circle is being swept out in the complex plane.