Transcript Physics 130 - University of North Dakota

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  Fnet = 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
Force always
points toward the
equilibrium
position.
FSpring = -k x
x is displacement
(compression or extension)
from equilibrium.
“Simple” harmonic motion only when the force is
proportional to the displacement, x, as in
Hooke’s law.
Equilibrium Position, Fnet = 0
m
k
greatest displacement right (x = A)
FS,B points left = Fnet = -kx
complete oscillation, # of seconds = T (Period)
greatest displacement left (x = A)
FS,B points right = Fnet = -kx
motion is symmetric, max displacement left =
max displacement right
Warning!!!! The
acceleration is not constant
so a  v/t !!!
k
m
vave  “mid-time” velocity!!!
Fnet does equal ma, however
greatest displacement up (x = A)
WE,B - FS,B points down = Fnet = -kx
measure x from equilibrium position
Equilibrium Position Fnet = 0
greatest displacement down (x = A)
FS,B - WE,B points up = Fnet = -kx
measure x from equilibrium position
motion is symmetric, max displacement up =
max displacement down
A
complete oscillation,
# of seconds = T (Period)
Fnet = -kx
Fnet = ma
What we find:
m
T  2
k
 2
f 1/ T
l
g
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) Yes
mass (m) Yes
amplitude (A) No
initial conditions No
(how we get it going)