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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: ¨ θ =- ω 2 sin θ. (1) If θ ( t ) oscillates with a small amplitude, we have sin θ ≈ θ , and the solution is simple: θ ( t ) = A sin( ωt + φ ), with the amplitude A 1. The period is then τ = 2 π ω . What if the amplitude is NOT small? Provided that the amptitude is less than π , the motion of θ is still periodic. How to find its period? The recipe follows. Multiplying both sides of Eq. (1) by ˙ θ , we find d dt ˙ θ 2 2 = d dt ω 2 cos θ . Therefore d dt ˙ θ 2 2- ω 2 cos θ = 0 . So ˙ θ 2 2- ω 2 cos θ = C is a constant. Physically, this is often associated with the conservation of energy. C is related to the amplitude of the oscillation: C =- ω 2 cos A. So ˙ θ = ± √ 2 ω (cos θ- cos A ) 1 / 2 , | θ | < A. The period is 1 τ = 2 Z A θ =- A dt = 2 Z A θ =- A dθ dθ dt = 2 Z A θ =- A dθ ˙ θ = 2 Z A- A dθ √ 2 ω (cos θ- cos A ) 1...
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This note was uploaded on 01/22/2012 for the course PHYS 3202 taught by Professor Tan during the Spring '11 term at Georgia Tech.
- Spring '11