Transcript Physics 131: Lecture 14 Notes

Physics 151: Lecture 33
Today’s Agenda

Topics
Potential energy and SHM
Resonance
Physics 151: Lecture 30, Pg 1
See text: 13.3
Energy in SHM

For both the spring and the pendulum, we can derive the
SHM solution using energy conservation.

The total energy (K + U) of a
system undergoing SMH will
always be constant!
U
K

This is not surprising since there
are only conservative forces
present, hence energy is conserved.
E
-A
0
U
A
s
Physics 151: Lecture 30, Pg 2
See text: Fig. 13.11
SHM and quadratic potentials



SHM will occur whenever the potential is quadratic.
Generally, this will not be the case:
For example, the potential between
H atoms in an H2 molecule looks
U
something like this:
K
E
U
-A
0
U
A
x
x
Physics 151: Lecture 30, Pg 3
See text: Fig. 13.11
SHM and quadratic potentials...
However, if we do a Taylor expansion of this function about the
minimum, we find that for small
displacements, the potential
IS quadratic:
U
U(x) = U(x0 ) + U(x0 ) (x- x0 )
1
+ U (x0 ) (x- x0 )2+....
2
U
x0
U(x) = 0 (since x0 is minimum
of potential)
x = x - x0 and U(x0 ) = 0
1
Then U(x) = U (x0 ) x 2
2
Define
x
x
Physics 151: Lecture 30, Pg 4
See text: Fig. 13.11
SHM and quadratic potentials...
1
U(x) = U (x0) x 2
2
Let k = U (x0)
U
U
x0
Then:
x
1
U(x) = k x 2
2
x
SHM potential !!
Physics 151: Lecture 30, Pg 5
What about Friction?



Friction causes the oscillations to get smaller over time
This is known as DAMPING.
As a model, we assume that the force due to friction is
dx
proportional to the velocity.
F f  b
dt
 F  ma
dx
LHS

F

F

F


kx

b

S
f
dt
d 2x
RHS  ma  m 2
dt
dx
d 2x
 kx  b  m 2
dt
dt
Physics 151: Lecture 30, Pg 6
What about Friction?
dx
d 2x
 kx  b  m 2
dt
dt
We can guess at a new solution.
x  A exp( 
bt
) cos(t   )
2m
With,
2
k  b 
b 

2


  o  

m  2m 
2
m


2
Physics 151: Lecture 30, Pg 7
What about Friction?
bt
x  A exp(  ) cos(t   )
2m
What does this function look like?
(You saw it in lab, it really works)
1.2
1
0.8
0.6
0.4
A
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
t
Physics 151: Lecture 30, Pg 8
What about Friction?
bt
x  A exp(  ) cos(t   )
2m
There is a cuter way to write this function if you
remember that exp(ix) = cos x + i sin x .
x  A exp(i (t   ))

2
k  b 
b


i

m  2m 
2m
   o2   2  i
Physics 151: Lecture 30, Pg 9
Damped Simple Harmonic Motion
  o2   2  i

Frequency is now a complex number! What gives?
Real part is the new (reduced) angular frequency
Imaginary part is exponential decay constant
o  
underdamped
o  
critically damped
o  
overdamped
Physics 151: Lecture 30, Pg 10
See text: 13.6
Driven SHM with Resistance

To replace the energy lost to friction, we can drive the
motion with a periodic force. (Examples soon).
F = F0 cos(t)

Adding this to our equation from last time gives,
ds
d 2s
F0 cos(t )  ks  b  m 2
dt
dt
Physics 151: Lecture 30, Pg 11
See text: 13.6
Driven SHM with Resistance

So we have the equation,
ds
d 2s
F0 cos(t )  ks  b  m 2
dt
dt

As before we use the same general form of solution,
s  A cos(t  )

Now we plug this into the above equation, do the
derivatives, and we find that the solution works as long as,
A
F0 / m
( 2   02 ) 2  (
b 2
)
m
Physics 151: Lecture 30, Pg 12
See text: 13.6
Driven SHM with Resistance




So this is what we need to think about the amplitude of the
oscillating motion,
F0 / m
A
b
( 2   02 ) 2  ( ) 2
m
Note, that A gets bigger if Fo does, and gets smaller if b or
m gets bigger. No surprise there.
Something more surprising happens if you drive the
pendulum at exactly the frequency it wants to go,  
k m
Then at least one of the terms in the denominator
vanishes and the amplitude gets real big. This is known as
resonance.
Physics 151: Lecture 30, Pg 13
See text: 13.6
Driven SHM with Resistance

Now, consider what b does,
F0 / m
A
b
( 2   02 ) 2  ( ) 2
m
b small
b middling
b large
  0
Physics 151: Lecture 30, Pg 14
Lecture 33, Act 1
Resonant Motion

Consider the following set of pendula all attached to the
same string
A
D
B
If I start bob D swinging which of the
others will have the largest swing amplitude ?
(A)
(B)
C
(C)
Physics 151: Lecture 30, Pg 15
Lecture 33, Act 1
Solution

The frequency of a pendulum is  

The driving frequency is  D 
g
L
g
LD

For each pendulum the natural frequency is  0 

D  0 for pendulum B so it is in resonance.
g
L
The answer is (B)
Physics 151: Lecture 30, Pg 16

Recap of today’s lecture
Chapter 13
Resonance
Physics 151: Lecture 30, Pg 17