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2/2/2016
Hooke’s Law
F=-kx
Simple Harmonic Motion
Physics 116: 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
rω2
Moving left
Moving Right
X=0
Completed cycle
acceleration
v 2
=
r
(r ω )2
r
1
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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.
•
•
•
•
•
l
ΣFon mass =Fspring-mg
ΣFon mass =-k(Δl)-mg
ΣFon mass =k(x+d)-mg
ΣFon mass =kx+kd-mg
ΣF=k(x)
d
x
http://en.wikipedia.org/wiki/Simple_harmonic_motion
Springs in SHM
• Tangential velocity
• Maximum energy
• A little algebra
The Circle and SHM
2πA
T
2
2
1
1
2 kA = 2 mvmax
vmax =
 2πA 
kA2 = m

 T 
T = 2π
• v = -vmax sin θ
• v = -Aω sin θ
• a = -amax cos θ
• a = -A ω2 cos θ
2
m
k
Equations of Motion
• x = xmax cos θ
• x = A cos θ
=
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
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Forces and acceleration for a cart on a
spring
• According to Hooke's law, the force that the
stretched spring exerts on a cart in the
x-direction is:
Period of vibrations of a cart attached
to a spring
• Starting with:
• And using:
• Using Newton's second law, we get:
• We get:
• The cart's acceleration ax is proportional to
the negative of its displacement x from the
equilibrium position.
© 2014 Pearson Education, Inc.
• In this expression for period, there is no
dependency on the amplitude.
© 2014 Pearson Education, Inc.
Pendulum (Fig 19-11)
Pendulum in SHM
• Ft = - m g sin θ
• Ft = - m g θ
=
=
s
Ft = - m g = − ks
L
mg
k=
L
m
T = 2π
T = 2π
Pendulum & Spring
=2
T = 2π
m
k
What were the assumptions for a pendulum in SHM?
mg
L
L
g
A Problem: Solve from equation
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)
 πt 
x = (0.30m) cos 
3
x = A cos θ
0.30 m (c) 1 6 Hz (d) 6.0 s
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A Problem: Solve for everything
A Problem: Solve for everything
• 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?
• 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?
Reading SHM
Graphs
Solve for everything
T = 2π
m
k
T = 2π
.05kg
= 2π (.0707s ) = 0.44s
10 Nm
f = 2.25Hz
A = .25m
x vs t
2.50
2.00
1.50
1.00
Things that are not constant:
ω=
Position
Times
θ
Mass x
Mass
velocity
Mass Accel
α=
vt=
0
t=
Look at the
graphs and find:
T=
f=
A=
ω=
α=
Vt =
Ar =
Things that are
constant:
ar=
12.5 cm
f=
T=
t=0.44 sec
A=
Reading SHM
Graphs
0.00
0.50
1.00
1.50
2.00
2.50
3.00
-1.00
-1.50
-2.00
-2.50
v vs t
8.00
6.00
4.00
2.00
0.00
-2.00
0.00
0.50
1.00
1.50
2.00
2.50
3.00
-4.00
-6.00
-8.00
Motion on a Spring
x vs t
2.50
0.50
-0.50 0.00
2.00
1.50
1.00
• At t=0.5 sec. find
for the shadow:
T, θ, x, v, a
• If k = 5 N/m, what
is the mass on
the spring?
0.50
Situation
0.00
-0.50 0.00
0.50
1.00
1.50
2.00
2.50
3.00
-1.00
-1.50
-2.00
-2.50
v vs t
8.00
6.00
4.00
2.00
0.00
-2.00
-4.00
-6.00
-8.00
0.00
0.50
1.00
1.50
2.00
2.50
3.00
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
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
(-)
(-) Max
Completed cycle
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Problem
Solve for everything
Reading SHM
Graphs
x vs t
2.50
2.00
1.50
1.00
Things that are
constant:
Things that are not constant:
ω=14.1 rad/sec
Position
Times
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
t=0.074 sec
t=0.44 sec
π/3
60°
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
Look at the graphs
and find:
0.50
0.00
-0.50 0.00
T = 2 sec
f = 0.5 sec
A=2m
ω = 3.14 Hz
α=0
vt = 6.28 m/s
ar = 19.7 m/s^2
-1.00
For the Shadow @
t=0.5 sec
8.00
T = 2 sec
θ = 1.57 rad
x = -1.97 m
v = -4.1 m/s
a = +14.9 m/s^2
For k=5 N/m, Mass = 0.51 kg
0.50
1.00
1.50
2.00
2.50
3.00
-1.50
-2.00
-2.50
v vs t
6.00
4.00
2.00
0.00
-2.00
0.00
0.50
1.00
1.50
2.00
2.50
3.00
-4.00
-6.00
-8.00
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