Physics 211: Lecture 24 Todays Agenda simple harmonic motion

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Physics 211 Lecture 24Todays Agendasimple
harmonic motion

• Introduction to Simple Harmonic Motion
• Horizontal spring mass (no gravity)
• The meaning of all these sines and cosines
• they are solutions to the differential equation
• Vertical spring mass (gravity)
• The energy approach
• The simple pendulum
• The rod pendulum

2
statics

• Forces
• Nothing is moving
• Nothing is set in motion
• Torques
• Nothing is rotating
• Nothing is induced to rotate

3
Lecture 21, Act 1 Solution

• Consider the torque about the hinge between the
  strut the wall

due to gravity
due to wire r x F, why 30 degrees?
Free body diagram Calculate forces and torques
T1
300
does not depend on length of massless beam! What
happens when q is small?
L
M
4
Hinged Beams...lecture 21

• What we want to find
• (Ax,Ay) and (Bx,By) these are forces acting on
  beams at A B.These keep the beams from moving
• What we know
• Sum of forces in x and y directions is zero
• Torque around any axis (A, B and C) is also zero
• WHY??
• No rotation! Its static

m2g
??
C
m1g
L
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Identify the forces Calculate force and torque
equations ( 0)
What about the cross product?
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Simple Harmonic Motion (SHM)
Horizontal Spring

• 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...

7
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

Our job is to find x(t)
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SHM Dynamics...
define
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SHM Dynamics...
Movie (shm)
Shadow

• But wait a minute...what does angular frequency ?
  have to do with moving back forth in a straight
  line ??

• y R cos ? R cos (?t)

y
1
1
1
2
2
3
3
?
0
x
4
-1
4
6
6
5
5
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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)

ok
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Derivation
We want to use the most general solution

• 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?
So we can use x A cos(?t ?) as the most
general solution! Still two parameters to
specify, amplitude and phase
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SHM Solution...

• Drawing of A cos(?t )
• A amplitude of oscillation

T 2?/?
A
?
??
?
??
-??
A
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SHM Solution...

• Drawing of A cos(?t ?)

?
?
??
?
??
-??
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SHM Solution...

• Drawing of A cos(?t - ?/2) A sin(wt)

? ??/2
A
?
??
?
??
-??
A sin(?t)!
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Meaning of the solution to SHM Solution...

• Drawing of A cos(?t ?)
• What does this really mean?
• Just that the amplitude and phase are both able
  to be set

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Lecture 24, Act 1Simple Harmonic Motion

• If you added the two sinusoidal waves shown in
  the top plot, what would the result look like?

(a)
(b)
(c)
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Lecture 24, Act 1 Solution

• Recall your trig identities

So
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
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What about Vertical Springs?
Vertical Spring

• 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
y 0
F -ky
Which has solution y A cos(?t ?)
where
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160
UNET UG US
140
U of Spring Gravity
120
U
100
US 1/2ky2
80
60
UG mgy
40
20
0
-10
-8
-6
-4
-2
0
2
4
6
8
10
-20
y
-40
-60
ye
0
shift due to mgy term
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What about Vertical Springs?Alternate treatment
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SHM So Far

• The most general solution is x A cos(?t ?)
• where A amplitude
• ? angular 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 net force is zero!

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The Simple Pendulum
Simple Pendulum

• A pendulum is made by suspending a mass m at the
  end of a string of length L. Find the angular
  frequency of oscillation for small
  displacements.

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Aside sin ? and cos ? for small ?

• A Taylor expansion of sin ? and cos ? about ? 0
  gives

and
So for ? ltlt 1,
and
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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 ? I????I??mL2

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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 gt T2 (c) T1 lt T2
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Lecture 24, Act 2 Solution

• We have shown that for a simple pendulum

Since

• If we make a pendulum shorter, it oscillates
  faster (smaller period)

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Lecture 24, Act 2 Solution
Standing up raises the CM of the swing, making it
shorter!
Since L1 gt L2 we see that T1 gt T2 .
L2
L1
T1
T2
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The Rod Pendulum

• A pendulum is made by suspending a thin rod of
  length L and mass m at one end. Find the angular
  frequency of oscillation for small displacements.

z
?
x
CM
L
mg
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The Rod Pendulum...

• The torque about the rotation (z) axis is ?
  -mgd -mg(L/2)sinq ? -mg(L/2)q for small q
• In this case
• So ? I???becomes

z
d
I
L/2
?
x
CM
L
d
mg
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Lecture 24, Act 3Period
Physical Pendulum

• What length do we make the simple pendulum so
  that it has the same period as the rod pendulum?

(a) (b) (c)
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Lecture 24, Act 3Solution
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Recap of todays 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 29, 45,
  65, 93