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viewnote3 - PHYS 302 Unit 3 Viewing Notes Athabasca...

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PHYS 302 : Unit 3 Viewing Notes – Athabasca University 1 PHYS 302: Vibrations and Waves - Unit 3 Viewing notes As in Unit 2, Lewin considers a mass on a spring, with b/m= γ and ω 0 2 = k/m . Initially, a force F 0 cos ω t is considered to be applied to the object. Summing the forces in Newton’s second law, 0 cos mx kx bx F t    ; and much like near the central, mathematically-oriented “18 minutes” of Unit 2, in the complex plane, reorganized, and with the RHS nonzero, one gets 0 2 0 F j t m z z z e  (3:30). We propose a solution ( ) i t z Ae , which represents, in the long term, an imposition of vibration at frequency ω . This is the steady state solution: although it changes in time because of the vibration, the overall state does not change; it just keeps vibrating. So steady state does not mean static . Doing the derivatives and substituting, 0 2 2 ( ) 0 ( ) F j t j t m j Ae e   (6:30). Equivalently, 0 0 2 2 0 ( ) (cos sin ) F F j m m j A e j   . This is really two equations, one for the real part, and one for the imaginary part, which both need to be true at the same time. Separating them, 0 2 2 0 ( ) cos F m A and 0 sin F m j A j   or 0 sin F m A  . Whenever we have equations like this with sin and cos, we always think of the identity 2 2 cos sin 1 and of the tangent. In squaring and adding these equations we have 0 2 2 2 2 2 2 0 ( ) ( ) cos F m A and 0 2 2 2 ( ) ( ) sin F m A  , giving 0 0 2 2 2 2 2 2 2 2 2 2 0 ( ) ( ) ( ) (cos sin ) ( ) F F m m A A  . Thus, 0 2 2 2 2 2 2 0 (( ) ( ) ) ( ) F m A  . Dividing through and taking the square root gives us 0 2 2 2 2 0 / ( ) ( ) F m A  as the amplitude of driven, damped motion (9:40). Although this function is complicated, it should be immediately clear that the amplitude will be highest when the driving frequency ω
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