VIEWS: 40 PAGES: 9 POSTED ON: 5/23/2012 Public Domain
Given a potential energy graph, oscillations will occur between turning points determined by E U xt E xt Even the most asymmetric well can be approximated by a parabola if the system stays close to the bottom. Simple Harmonic Motion k and x need not refer to 1 2 U kx spring parameters, but the spring-mass system is an 2 important example 2 dU d x F kx kx ma m 2 dx dt 2 d x k 2 x0 Standard form of harmonic oscillator equation dt m angular frequency k Solution: xt A cost 2 m amplitude phase phase angle A and determined by initial conditions vt A sin t Exercise: A spring-mass with k = 25 N/m and m = 5 kg is pulled out 2 m and released. Find its position at any later time. 0 x0 A cos 2 v0 A sin 0 A2 xt 2 cos 5t The motion will have completed one full cycle when the phase has increased by 2π. This time interval is called the period of the oscillation. t T t 2 2 T Energy Considerations How can the total energy be constant when both K and U depend on time? 1 1 2 E mv kx2 2 2 E m A sin t k A cos t 2 1 2 1 2 2 1 2 E kA sin 2 t cos 2 t 2 1 2 E kA 2 Vertical Springs When a spring mass-system k vibrates vertically, an extra force, gravity is involved. How does this affect the motion? kx x m m netF mg kx mg d 2x dt 2 k xg m ? Rewrite: d 2x 2 k x m mg 0 k dt mg d 2 x d 2 x Define new displacement: x x Note that: 2 k 2 dt dt d 2 x k Then we have: 2 x 0 Gravity does not change the period. dt m All effects of gravity can be “ignored” if we choose to measure “X” from the new equilibrium point. Example: A 1 kg mass hangs vertically from a spring with k = 20 N/m. A .1 kg dart moving at 20 m/s is fired into the mass and sticks into it. Find the period and amplitude of the resulting harmonic motion. The old equilibrium point is The new equilibrium point is 110 1.110 i x0 .50 m P0 Pf x0f .55 m 20 .120 1.1V 20 1kg m V 1.82 s The period just depends on k and the mass oscillating: m 1.1 T 2 2 1.47 s k 20 Use energy conservation with “x” referred to new equilibrium point: E0 E f 1 1.11.82 20.05 20A2 2 1 2 1 A .43m 2 2 2 Other Oscillators Torsion Pendulum As wire twists, a “restoring torque is exerted. Fixed axis rotation k I I d 2 d 2 k I k 0 dt 2 dt 2 I t m cost I T 2 k Simple Pendulum T L mg I mL 2 mg sin L I d 2 g sin 0 dt 2 L ? Small angle approximation: sin d 2 g T 2 L 0 dt 2 L g Compound Pendulum H mg sin H I CM About pivot d 2 mgH 2 sin 0 dt I d 2 mgH Small angle approximation: 2 0 sin dt I I T 2 mgH