Class 14: Simple harmonic motion Class 14: Simple harmonic motion - PowerPoint PPT Presentation

Class 14: Simple harmonic motion Class 14: Simple harmonic motion

Origin of simple harmonic motion U(x) = ½ kx 2 Total energy Total energy F = - kx x V=0 a is max V=0, a is max. V=0 a is max V=0, a is max. a=0, v is max.

Equation of motion Equation of motion = & & m m x x - k x k x

Solution = ω + φ x A cos ( t ) 0 or ω ω + + φ φ = i( i( t t ) ) x A e 0 k k Natural frquency or ω 0 = resonance frequency m A and φ to be determined by initial conditions. A is the amplitude of oscillation.

Old slide Circular motion with uniform speed = a 0 T = v constant v 2 v = = ω 2 a r a R a R R r r θ r r 1 1 = f T ω ω = = π π 2 2 f f π 2 r = ω = v r T T

Conservation of energy Conservation of energy At all points in the path, total energy = constant 1 1 1 + = 2 2 2 mv kx kA 2 2 2 2 2 2 1 1 1 + = 2 2 2 mv kx mv max 2 2 2 2 2 2