11/18 Simple Harmonic Motion  HW “Simple Harmonic Motion” Due Thursday 11/21  Exam 4 Thursday 12/5 Simple Harmonic Motion Angular Acceleration and Torque.

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Presentation transcript:

11/18 Simple Harmonic Motion  HW “Simple Harmonic Motion” Due Thursday 11/21  Exam 4 Thursday 12/5 Simple Harmonic Motion Angular Acceleration and Torque Angular Momentum I’m out of town Monday 12/2 (after Thanksgiving)

Simple Harmonic Motion (SHM)  Real objects are “elastic” (to some extent)  Elastic:  Object returns to original shape after deformation  Original shape called “Equilibrium”  Object resists deformation with force  Force called “Restoring Force”  Force proportional to deformation  Force is in opposite direction to deformation Force points toward equilibrium, deformation points away from equilibrium

Simple Harmonic Motion (SHM)  Definitions:  Displacement  x  “deformation from equilibrium”  Equilibrium  F net = 0 as usual  Oscillation  one complete cycle  Period T  time for one complete oscillation  Frequency f  # of oscillations in one second example: 1.65Hz (in units of “Hertz”)  Amplitude A  maximum displacement from equilibrium

Hooke’s Law = Simple Harmonic Motion F Spring = -k  x “Simple” harmonic motion only when the force is proportional to the displacement,  x, as in Hooke’s law.  x is displacement (compression or extension) from equilibrium. Force always points toward the equilibrium position.

m k Equilibrium Position, F net = 0 greatest displacement right (  x = A) F S,B points left = F net = -k  x greatest displacement left (  x = A) F S,B points right = F net = -k  x complete oscillation, # of seconds = T (Period) motion is symmetric, max displacement left = max displacement right

m k motion is symmetric, max displacement up = max displacement down Equilibrium Position F net = 0 greatest displacement down (  x = A) F S,B - W E,B points up = F net = -k  x measure  x from equilibrium position greatest displacement up (  x = A) W E,B - F S,B points down = F net = -k  x measure  x from equilibrium position complete oscillation, # of seconds = T (Period) A Warning!!!! The acceleration is not constant so a   v/  t !!! v ave  “mid-time” velocity!!! F net does equal ma, however

F net = ma F net = -k  x

What we find:

A block hangs from a spring and is pulled down 10 cm and released. It bounces up and down at a rate of 3 times every second. How could this rate be increased? You may choose any of:  increase the mass  decrease the mass  push it faster  make the spring stiffer  make the spring less stiff  pull it down farther  don’t pull it down as far  other?

What matters? Period (or frequency) affected by:  spring constant (k)  mass (m)  amplitude (A)  initial conditions (how we get it going) Yes No