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Unformatted text preview: Finding the period of some oscillation Suppose that we obtain the following equation of motion for a physical system: =- 2 sin . (1) If ( t ) oscillates with a small amplitude, we have sin , and the solution is simple: ( t ) = A sin( t + ), with the amplitude A 1. The period is then = 2 . What if the amplitude is NOT small? Provided that the amptitude is less than , the motion of is still periodic. How to find its period? The recipe follows. Multiplying both sides of Eq. (1) by , we find d dt 2 2 = d dt 2 cos . Therefore d dt 2 2- 2 cos = 0 . So 2 2- 2 cos = C is a constant. Physically, this is often associated with the conservation of energy. C is related to the amplitude of the oscillation: C =- 2 cos A. So = 2 (cos - cos A ) 1 / 2 , | | < A. The period is 1 = 2 Z A =- A dt = 2 Z A =- A d d dt = 2 Z A =- A d = 2 Z A- A d 2 (cos - cos A ) 1...
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- Spring '11