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 kx2 Total energy Total energy

kx

• F=

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

k x

• x

m = & & k x x m

Solution

) t ( cos A x φ ω + =

) t i(

• r

φ ω + ) t i(

e A x

φ ω +

=

k m k

0 =

ω

Natural frquency or resonance frequency

A and φ to be determined by initial

• conditions. A is the amplitude of oscillation.

Circular motion with uniform speed Old slide

T

constant v a = =

aR

2 2 R

r r v a ω = =

v aR r θ

r 1

r

f 2 T 1 f π ω = = T r 2 r v f 2 π ω π ω = = = T

Conservation of energy Conservation of energy At all points in the path, total energy = constant

2 2 2

kA 2 1 kx 2 1 mv 2 1 = + 2 2 2

2 max 2 2

mv 2 1 kx 2 1 mv 2 1 = + 2 2 2