Transcript Physics 201: Lecture 1

Physics 201:
Chapter 14 – Oscillations
Dec 8, 2009
7/18/2015

Hooke’s Law

Spring

Simple Harmonic Motion

Energy
Physics 201, UW-Madison
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Springs

Hooke’s Law: The force exerted by a spring is proportional
to the distance the spring is stretched or compressed from
its relaxed position.
FX = -k x
Where x is the displacement from
the relaxed position and k is the
constant of proportionality.
(often called “spring constant”)
relaxed position
FX = -kx > 0
FX = 0
x0
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FX = - kx < 0
xx
x=0 x > 0
x=0
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Simple Harmonic Motion (SHM)


We know that if we stretch
a spring with a mass on
the end and let it go, the
mass will oscillate back
and forth (if there is no
friction).
This oscillation is called
Simple Harmonic
Motion, and is actually
easy to understand...
k
k
k
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m
m
m
3
SHM Dynamics


At any given instant we know
that F = ma must be true.
F = -kx
k
But in this case F = -kx
a
m
and ma =

So: -kx = ma =
d2x
m 2
dt
x
d2x
k
 x
2
dt
m
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a differential equation
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SHM Dynamics...
d2x
k
 x
2
dt
m
define
d2x
2



x
2
dt

k
m
The angular frequency: ω
The peroid: T = 2π / ω
Try the solution x = A cos( t)
dx
  Asin  t 
dt
d2x
2
2



A
cos

t



x


2
dt
This works, so it must be a solution!
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SHM Dynamics...

We just showed that
d2x
2



x
2
dt
(from F = ma)
has the solution x = A cos( t) but … x = A sin( t) is also a solution.

The most general solution is a linear combination (sum) of these two
solutions!
x  Bsin t   C cos t   Acos  t   
dx
  B cos  t    C sin  t 
dt
d2x
2
2
2



Bsin

t


C
cos

t



x ok




2
dt
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SHM Dynamics...
x  Acos  t   
The period: T = 2π / ω
But wait a minute...what does angular frequency  have to
do with moving back & forth in a straight line ??

y
1
1
2
3
x
4
6
5
0
-1
1
2

3

2
4

6
5
The spring’s motion with amplitude A is identical to the
x-component of a particle in uniform circular motion with
radius A.
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Example

A mass m = 2 kg on a spring oscillates with amplitude A = 10 cm. At t = 0
its speed is maximum, and is v = +2 m/s.
What is the angular frequency of oscillation  ?
What is the spring constant k?
vMAX 2 m s


 20 s 1
A
10 cm
vMAX = ωA
Also:

k
m
k = m 2
So k = (2 kg) x (20 s -1) 2 = 800 kg/s2 = 800 N/m
k
m
x
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Initial Conditions
Use “initial conditions” to determine phase  ! x(t) = A cos( t +  )
v(t) = - A sin( t +  )
Suppose we are told x(0) = 0 , and x is
a(t) = - 2A cos( t +  )
initially increasing (i.e. v(0) = positive):
 = /2 or -/2
 <0
x(0) = 0 = A cos( )
v(0) > 0 = - A sin( )
So  = -/2

k
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cosθ sinθ
m
0

2
x
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Initial Conditions...
So we find  = -/2
x(t) = A cos( t - /2 )
v(t) = - A sin( t - /2 )
a(t) = - 2A cos( t - /2 )
x(t) = A sin( t)
v(t) =  A cos( t)
a(t) = - 2A sin( t)
A

k
m
0
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x(t)
t
2
-A
x
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What about Vertical Springs?

We already know that for a vertical spring
if y is measured from
the equilibrium position
j

So this will be just like the horizontal case:
2
d y
k y  ma  m 2
dt
k
y=0
Which has solution y = A cos( t +  )
where
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
k
m
Physics 201, UW-Madison
m
F = -ky
11
The Simple Pendulum...

Recall that the torque due to gravity about the rotation (z) axis is
 = -mgd
d = Lsin  L for small 
so  = -mg L
But
d 2
I 2
dt
z
 = I  where I = mL2

d 2
2




2
dt
where

g
L
Differential equation for simple harmonic motion!
  0 cos t   
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L
m
d
mg
12
Simple Harmonic Motion:
Summary
k

2
Force:
d s
2
  s
2
dt
s
k
m
0
m
k
m
s
0
Solution:
s = A cos( t +  )
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
Physics 201, UW-Madison
g
L
s
L
13
The potential energy function
1 2
U  kx
2
1 2
U  kx
2
1 2
K  mv
2
E U  K
1 2 1 2
 kA  mv max
2
2
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Energy for simple harmonic motion: E=K+U
x  Acos t 
t
U
1 2 1 2
U  kx  kA cos 2 t 
2
2
t
K
t
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dx
  Asin  t 
dt
1 2 1
K  mv  m 2 A2 sin 2  t 
2
2
v
E  K  U  K Max 
1
1
m 2 A 2  U Max  kA 2
2
2 15
Energy conservation
m
x
x=0
Etotal = 1/2 Mv2 + 1/2 kx2 = constant
KE
PE
KEmax = Mv2max /2 = Mω2A2 /2 =kA2 /2
PEmax = kA2 /2
Etotal = kA2 /2
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