physics 111: lecture 24, pg 1 physics 111: lecture 24 today’s agenda l introduction to simple...
TRANSCRIPT
Physics 111: Lecture 24, Pg 1
Physics 111: Lecture 24
Today’s Agenda
Introduction to Simple Harmonic Motion Horizontal spring & mass
The meaning of all these sines and cosines Vertical spring & mass The energy approach The simple pendulum The rod pendulum
Physics 111: Lecture 24, Pg 2
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 very easy to understand...
km
km
km
Horizontal
Spring
Physics 111: Lecture 24, Pg 3
SHM Dynamics
At any given instant we know that F = ma must be true.
But in this case F = -kx and ma =
So: -kx = ma =
k
x
m
F = -kx
a
md x
dt
2
2
d x
dt
k
mx
2
2 a differential equation for x(t)!
md x
dt
2
2
Physics 111: Lecture 24, Pg 4
SHM Dynamics...
d x
dt
km
x2
2
d x
dtx
2
22
km
Try the solution x = A cos(t)
tsinAdtdx
xtcosAdt
xd 222
2
This works, so it must be a solution!
define
Where w is the angular
frequency of motion
Physics 111: Lecture 24, Pg 5
SHM Dynamics...
y = R cos = R cos (t)
But wait a minute...what does angular frequency have to do with moving back & forth in a straight line ??
Movie (shm)
x
y
-1
1
0
1 12 2
3 3
4 45 5
6 62
Shadow
Physics 111: Lecture 24, Pg 6
SHM Solution
We just showed that (which came from F = ma)
has the solution x = A cos(t) .
This is not a unique solution, though. x = A sin(t) is also a solution.
The most general solution is a linear combination of these two solutions!
x = B sin(t)+ C cos(t)
d x
dtx
2
22
tsinCtcosBdtdx
xtcosCtsinBdt
xd 2222
2
ok
Physics 111: Lecture 24, Pg 7
Derivation:
x = A cos(t + ) is equivalent to x = B sin(t)+ C cos(t)
x = A cos(t + )
= A cos(t) cos - A sin(t) sin
where C = A cos() and B = A sin()
It works!
= C cos(t) + B sin(t)
We want to use the most general solution:
So we can use x = A cos(t + ) as the most general solution!
Physics 111: Lecture 24, Pg 8
SHM Solution...
Drawing of A cos(t ) A = amplitude of oscillation
T = 2/
A
A-
Physics 111: Lecture 24, Pg 9
SHM Solution...
Drawing of A cos(t + )
-
Physics 111: Lecture 24, Pg 10
SHM Solution...
Drawing of A cos(t - /2)
A
= /2
= A sin(t)!
-
Physics 111: Lecture 24, Pg 11
Lecture 24, Act 1Simple Harmonic Motion
If you added the two sinusoidal waves shown in the top plot, what would the result look like?
-0.60
-0.40
-0.20
0.00
0.20
0.40
0.60
0
100
200
300
400
500
600
700
800
900
1000
-1.00
-0.50
0.00
0.50
1.000
100
200
300
400
500
600
700
800
900
1000
(a)
-2.00
-1.50
-1.00
-0.50
0.00
0.50
1.00
1.50
2.00
0
100
200
300
400
500
600
700
800
900
1000(b)
(c)-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0
100
200
300
400
500
600
700
800
900
1000
Physics 111: Lecture 24, Pg 12
Lecture 24, Act 1 Solution
Recall your trig identities:
2
BA
2
BA2BA coscoscoscos
tt coscos
22a
cos
2b
bta cosSo
Where
The sum of two or more sines or cosines having the same frequency is just another sine or cosine with the same frequency.
The answer is (b).
Prove this with Excel
Physics 111: Lecture 24, Pg 13
What about Vertical Springs?
We already know that for a vertical spring if y is measured from
the equilibrium position
The force of the spring is the negative derivative of this function:
So this will be just like the horizontal case:
-ky = ma =
j
k
mF = -ky
y = 0
U ky1
22
kydydU
F
md y
dt
2
2
Which has solution y = A cos(t + )
km
where
Vertical
Spring
Physics 111: Lecture 24, Pg 14
SHM So Far
The most general solution is x = A cos(t + ) where A = amplitude
= frequency = phase
For a mass on a spring
The frequency does not depend on the amplitude!!!We will see that this is true of all simple harmonic
motion! The oscillation occurs around the equilibrium point where
the force is zero!
km
Physics 111: Lecture 24, Pg 15
The Simple Pendulum
A pendulum is made by suspending a mass m at the end of a string of length L. Find the frequency of oscillation for small displacements.
L
m
mg
z
Simple
Pendulum
Physics 111: Lecture 24, Pg 16
Aside: sin and cos for small
A Taylor expansion of sin and cos about = 0 gives:
...!5!3
sin53
...!4!2
1cos42
and
So for << 1, sin 1cos and
Physics 111: Lecture 24, Pg 17
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 = II=mL2 L
dm
mg
z
mgL mLd
dt 2
2
2
d
dt
2
22 g
Lwhere
Differential equation for simple harmonic motion!
= 0 cos(t + )
Physics 111: Lecture 24, Pg 18
Lecture 24, Act 2Simple Harmonic Motion
You are sitting on a swing. A friend gives you a small push and you start swinging back & forth with period T1.
Suppose you were standing on the swing rather than sitting. When given a small push you start swinging back & forth with period T2.
Which of the following is true:
(a) T1 = T2
(b) T1 > T2
(c) T1 < T2
Physics 111: Lecture 24, Pg 19
Lecture 24, Act 2 Solution
We have shown that for a simple pendulum gL
TLg
2Since T 2
If we make a pendulum shorter, it oscillates faster (smaller period)
Physics 111: Lecture 24, Pg 20
Lecture 24, Act 2 Solution
L1
L2
Standing up raises the CM of the swing, making it shorter!
T1 T2
Since L1 > L2 we see that T1 > T2 .
Physics 111: Lecture 24, Pg 21
The Rod Pendulum
A pendulum is made by suspending a thin rod of length L and mass m at one end. Find the frequency of oscillation for small displacements.
Lmg
z
xCM
Physics 111: Lecture 24, Pg 22
The Rod Pendulum...
The torque about the rotation (z) axis is
= -mgd = -mg(L/2)sin q -mg(L/2) q for small q
In this case
So = Ibecomes
Ldmg
z
L/2
xCM
I 1
32mL
2
22
dt
dmL
3
1
2
Lmg
d
dt
2
22 3
2gL
where
d I
Physics 111: Lecture 24, Pg 23
Lecture 24, Act 3Period
(a) (b) (c)
What length do we make the simple pendulum so that it has the same period as the rod pendulum?
LR
LS
RS L32
L RS L23
L RS LL
Physical
Pendulum
Physics 111: Lecture 24, Pg 24
LR
LS
S = P if RS L32
L
Lecture 24, Act 3Solution
SS L
g
RR L2
g3
Physics 111: Lecture 24, Pg 25
Recap of today’s lecture
Introduction to Simple Harmonic Motion (Text: 14-1)Horizontal spring & mass
The meaning of all these sines and cosines Vertical spring & mass (Text: 14-3) The energy approach (Text: 14-
2) The simple pendulum (Text: 14-3) The rod pendulum
Look at textbook problems Chapter 14: # 1, 13, 33, 55, 93