Download Vibrations and Waves

1/3/2012
Hooke’s Law
F=-kx
Simple Harmonic Motion
Physics 123: Eyres
Hooke’s Law
Springs and Kinematics
F=-kx
•
•
•
•
• F is the Force on ________ caused by
________
• X is really the ____________
• K is ____________
• The negative sign means_____________
F is the Force on Object caused by Spring
X is really the Spring Extension/Compression
K is Spring Constant
The negative sign means that
extension/compression are in the opposite
direction to F.
• Let’s Review the motion
of the object at
different locations on
an oscillating spring.
• Find the sign of the
– Position
– Velocity
– Acceleration
http://en.wikipedia.org/wiki/File:Muelle.gif
Motion on a Spring
Review the Quantities
Situation
Position (+ to Right) velocity
Pulled to right and
released
X=+A
Component
Symbol
Relation to
Rotational
Quantity
It comes from this:
X=0
Tangential
Displacement
s
θr
Circumference
= 2πr
Tangential
Velocity
vt
ωr
2πr ∆θ
=
r
Τ
∆t
Tangential
Acceleration
at
αr
look for the
pattern
Radial (Centripetal)
Acceleration
ar or ac
rω
ω2
Moving left
Moving Right
X=0
Completed cycle
acceleration
v 2
=
r
(r ω )2
r
1
1/3/2012
SHM: Hanging Spring
SHM: Hanging Spring
• The restoring force
must be proportional to
the displacement, or
approximately so.
• Note: When the spring
is stretched with a mass
in equilibrium it is
stretched by d.
Any system that undergoes
simple harmonic motion:
• When the system is displaced
from equilibrium there must exist
a restoring force that tends to
restore it to equilibrium.
• The restoring force must be
proportional to the displacement,
or approximately so.
•
•
•
•
•
ΣFon mass =Fspring-mg
ΣFon mass =-k(Δl)-mg
ΣFon mass =k(x+d)-mg
ΣFon mass =kx+kd-mg
ΣF=k(x)
http://en.wikipedia.org/wiki/Simple_harmonic_motion
Springs in SHM
• Tangential velocity
• Maximum energy
• A little algebra
2πA
T
2
2
1
1
2 kA = 2 mvmax
vmax =
 2πA 
kA = m

 T 
2
m
T = 2π
k
Equations of Motion
• x = xmax cos θ
• x = A cos θ
• v = -vmax sin θ
• v = -Aω sin θ
• a = -amax cos θ
• a = -A ω2 cos θ
Pendulum in SHM
2
• Ft = - m g sin θ
• Ft = - m g θ=-mgs
Ft = - m g
k=
s
= − ks
L
mg
L
T = 2π
T = 2π
m
mg
L
L
g
T = 2π
m
k
Equations of Motion
• What are the assumptions for which these
equations can be used?
• What if you have a different situation?
x=A cos (2πƒt) = A cos ωt
v = -2πƒA sin (2πƒt) = -A ω sin ωt
a = -4π2ƒ2A cos (2πƒt) = -Aω2 cos ωt
2
1/3/2012
A Problem
The motion of an object is
described by the
equation
Find
• (a) the position at t = 0
and t = 0.60 s,
• (b) the amplitude
• (c) the frequency
• (d) the period
(a)
0.30 m, 0.24 m (b)
A Problem
 πt 
x = (0.30m) cos 
3
x = A cos θ
• A 50.0-g object is
attached to a horizontal
spring with a force
constant of 10.0 N/m
and released from rest
with an amplitude of
25.0 cm. What is the
velocity of the object
when it is halfway to
the equilibrium position
if the surface is
frictionless?
Things that are
constant:
Things that are not constant:
ω=
Position
θ
Mass x
Mass
velocity
Mass Accel
α=
vt=
0
t=
m
k
T = 2π
.05kg
= 2π (.0707 s) = 0.44s
10 Nm
f = .71Hz
A = .25m
0.30 m (c) 1 6 Hz (d) 6.0 s
Solve for everything
Times
T = 2π
ar=
12.5 cm
f=
Reading SHM Graphs
• SHM_graphs.xlsx
• Look at the graphs and find:
T, f, A, ω, α, vt, ar
• At a given time find for the shadow:
T, θ, x, v, a
• If k = 5 N/m, what is the mass on the spring?
T=
t=0.44 sec
A=
Problem #16
Solve for everything
Motion on a Spring
Situation
Position (+ to Right) velocity
acceleration
Pulled to right and X=+A
released (speed up)
0
Max to left (-)
Moving left speed
up
Left (-)
(-)
0<x<+A
ω=14.13 rad/sec
Position
Times
Move L, at center
X=0
Left (-) Max
0
Move L (slows)
-A<x<0
(-)
(+)
Far to Left change
direction
-A
0
Max (+)
Moving Right speed -A<x<0
up
(+)
(+)
Move R at center
X=0
(+) Max
0
Move R slows
0<x<+A
(+)
(-)
At far right
X=+A
Things that are
constant:
Things that are not constant:
t=0
θ
Mass x
0
25 cm
Mass
velocity
0
Mass Accel
-50 m/s2
α=0
vt=3.53m/s
ar=50 m/s2
(-) Max
t=0.074 sec
π/3
60°
t=0.44 sec
2π
360°
12.5 cm -3.06 m/s
-25 m/s2
F=2.25Hz
T=0.44 s
25 cm
0
-50 m/s2
A=25 cm
Completed cycle
3