Transcript courses:lecture:oslec:os_energydiagram_wiki.ppt

PH421: Oscillations – lecture 1
Reading:
Taylor 4.6
(Thornton and Marion 2.6)
Giancoli 8.9
1
Goals for the pendulum module:
(1) CALCULATE the period of oscillation if we know the
potential energy (using exact and approximate forms);
specific example is the pendulum
(2) MEASURE the period of oscillation as a function of
oscillation amplitude
(3) COMPARE the measured period to models that make
different assumptions about the potential
(4) PRESENT the data and a discussion of the models in
a coherent form consistent with the norms in physics
writing
(5) CALCULATE the (approximate) motion of a
pendulum by solving Newton's F=ma equation
2
How do you calculate how long it takes to get from
one point to another?
Dx
Dt =
v
But what if v is not constant?
dx
= v(x)
dt
t finish
ò
t start
dx
dt =
v(x)
Separation of x and t variables!
x(t =t finish )
dx
dt = ò
v(x)
x(t =t start )
3
The case of a conservative force
Suppose total energy is CONSTANT
(we have to know it, or be able to
find out what it is)
E = K(x) + U(x)
2
dx
2
é dx ù
K(x) = m ê ú Þ
=±
[ E - U(x)]
dt
m
ë dt û
1
2
dt =
dx
2
±
[ E -U(x)]
m
4
Example: U(x) = ½ kx2 , the harmonic oscillator
U(x)
U(x) = kx
1
2
Energy
E
K(x) = mx KE
1
2
B
position
x0
Classical turning points
PE
dU
F(x) = = -kx
dx
K(x)
xL
2
2
xR
dx
xº
dt
5
Symmetry - time to go there is
the same as time to go back (no
damping)
Energy
E
xR
T = 2ò
xL
xL
x0
position
xR
dx
2
E - U(x)]
[
m
SHO - symmetry about x0
xL -> x0 same time as for x0 -> xR
xR
T = 4ò
x0
dx
2
éë E - 12 kx 2 ùû
m
6
Energy
SHO - do we get what we expect?
E
x= A
T =4
dx
ò
2
éë E - 12 kx 2 ùû
m
x=0
xL=-A
x0x=0
0
position
xR=A
Another way to specify E
is via the amplitude A
x
u=
A
E = 12 kA2
u =1
T =4
ò
u=0
http://www.wellesley.edu/Physics/
Yhu/Animations/sho.html
du
k
éë1 - u 2 ùû
m
Independent of A!
7
Energy
u =1
T =4
E
du
ò
k
éë1 - u 2 ùû
m
u=0
u = sin f
xL=-A
x0
position
xR=A
du = cos f df
1 - u = cos f
2
TSHO
m 2p
= 2p
º
k w0
You have seen this before
in intro PH, but you didn't
derive it this way.
j =p /2
T =4
òf
=0
cos f df
k
cos f
m
8
TSHO =
2p
w0
Period of SHO is INDEPENDENT OF AMPLITUDE
Why is this surprising or interesting?
As A increases, the distance and velocity change. How does this
affect the period for ANY potential?
A increases -> further to travel -> distance increases
-> period increases
A increases -> more energy -> velocity increases
-> period decreases
Which one wins, or is it a tie?
9
Look at example of simple pendulum
(point mass on massless string). This is
still a 1-dimensional problem in the
sense that the motion is specified by one
variable, q
K (q ) + U (q ) = E Total E
K (q ) = Iq KE
U (q ) = mgL(1- cosq ) PE
1
2
2
Your lab example is a plane pendulum. You will have to generalize:
10
what length does L represent in this case?
E = K (q ) + U (q ) Tot E
K (q ) = Iq KE
U (q ) = mgL(1- cosq ) PE
2
1
2
cosq » 1 - 2!1 q 2 + 4!1 q 4 +
(
U approx = mgL q
1
2
2
)
U exact = mgL (1 - cosq )
Plot these to compare to SHO to pendulum……
11
q
2
Energy
Compare 1- cosq with
1
2!
position
Period increases because v(x) decreases (E decreases) for the
same amplitude. Effect is magnified for larger amplitudes.
12
dx
dt =
=±
v
dx
xR
T = 2ò
2
E - U(x)]
[
m
xL
dx
2
E - U(x)]
[
m
Equivalent angular version?
dt =
dq
w
=
dq
q
K(q ) = 12 Iq 2
dt =
q=
2
I
( E - U(q ))
dq
2
I
( E - U(q ))
13
dt =
dq
2
I
( E - U(q ))
Integrate both sides
E and U(q) are known - put them in
Resulting integral
• do approximately by hand using series expansion
(pendulum period worksheet on web page)
OR
• do numerically with Mathematica (notebook on
14
web page)
Eqm:
E
Energy
dU
dx
dU
If
dx 2
=0
x0
2
K(x)
=k
x0
B
xL
position
x0
“Everything is a SHO!”
xR
U(x) = U(x0 ) + k(x - x0 )
1
2
2
(x - x0 )2 d 2U
( x - x0 )2 d 3U
dU
U ( x) = U ( x 0 ) + ( x - x 0 )
+
+
2
2!
3!
dx x0
dx x
dx 3 15
x0
0