Physics 111: Lecture 25 Todays Agenda

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Physics 111 Lecture 25Todays Agenda

• Recap of last lecture
• Using initial conditions to solve problems
• The general physical pendulum
• The torsion pendulum
• Energy in SHM
• Atomic Vibrations
• Problem Vertical Spring
• Problem Transport Tunnel
• SHM Review

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SHM and Springs
Force
Solution s A cos(?t ?)
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Velocity and Acceleration
Position x(t) A cos(?t ?) Velocity v(t)
-?A sin(?t ?) Acceleration a(t) -?2A
cos(?t ?)
by taking derivatives, since
xMAX A vMAX ?A aMAX ?2A
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Lecture 25, Act 1Simple Harmonic Motion

• A mass oscillates up down on a spring. Its
  position as a function of time is shown below.
  At which of the points shown does the mass have
  positive velocity and negative acceleration?

y(t)
(a)
(c)
t
(b)
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Lecture 25, Act 1 Solution

• The slope of y(t) tells us the sign of the
  velocity since

• y(t) and a(t) have the opposite sign since a(t)
  -w2 y(t)

a lt 0v gt 0
a lt 0v lt 0
y(t)
(a)
(c)
t
(b)
a gt 0v gt 0
The answer is (c).
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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 ?A
?
Also
k m?2
So k (2 kg) x (20 s -1) 2 800 kg/s2 800 N/m
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Initial Conditions
Use initial conditions to determine phase ?!
Suppose we are told x(0) 0 , and x is
initially increasing (i.e. v(0) positive)
x(0) 0 A cos(?) ? ?/2 or -?/2 v(0) gt 0
-?A sin(?) ? lt 0
? -?/2
So
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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)
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Lecture 25, Act 2Initial Conditions

• A mass hanging from a vertical spring is lifted a
  distance d above equilibrium and released at t
  0. Which of the following describes its velocity
  and acceleration as a function of time?

(a) v(t) -vmax sin(wt) a(t) -amax
cos(wt)
k
y
(b) v(t) vmax sin(wt) a(t) amax
cos(wt)
d
t 0
(c) v(t) vmax cos(wt) a(t) -amax
cos(wt)
0
(both vmax and amax are positive numbers)
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Lecture 25, Act 2 Solution
Since we start with the maximum
possibledisplacement at t 0 we know that y
d cos(wt)
k
y
d
t 0
0
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Review of Simple Pendulum

• Using ? I? and sin ? ? ? for small ?

?
?
I
We found
where
Which has SHM solution ? ?0 cos(?t ?)
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Review of Rod Pendulum

• Using ? I? and sin? ? ? for small ?

?
?
I
We found
where
Which has SHM solution ? ?0 cos(?t ?)
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General Physical Pendulum
Physical Pendulum

• Suppose we have some arbitrarily shaped solid of
  mass M hung on a fixed axis, and that we know
  where the CM is located and what the moment of
  inertia I about the axis is.
• The torque about the rotation (z) axis for small
  ? is (sin ? ? )
  
  
  ? -Mgd -MgR??

z-axis
R
?
x
CM
?
?
d
Mg
where
? ?0 cos(?t ?)
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Lecture 25, Act 3Physical Pendulum

• A pendulum is made by hanging a thin hoola-hoop
  of diameter D on a small nail.
• What is the angular frequency of oscillation of
  the hoop for small displacements? (ICM mR2 for
  a hoop)

pivot (nail)
(a) (b) (c)
D
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Lecture 25, Act 3 Solution
Hoop Pendulum

• The angular frequency of oscillation of the hoop
  for small displacements will be given by

(see Lecture 25 notes)
Use parallel axis theorem I Icm mR2
mR2 mR2 2mR2
pivot (nail)
cm x
R
m
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Torsion Pendulum

• Consider an object suspended by a wire attached
  at its CM. The wire defines the rotation axis,
  and the moment of inertia I about this axis is
  known.
• The wire acts like a rotational spring.
• When the object is rotated, the wire is twisted.
  This produces a torque that opposes the rotation.
• In analogy with a spring, the torque produced is
  proportional to the displacement ? -k?

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Torsion Pendulum...
Torsion Pendulum

• Since ? -k??? ? I???becomes

where
This is similar to the mass on spring except I
has taken the place of m (no surprise).
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Energy in SHM

• For both the spring and the pendulum, we can
  derive the SHM solution by using energy
  conservation.
• The total energy (K U) of a system undergoing
  SHM will always be constant!
• This is not surprising since there are only
  conservative forces present, hence KU energy is
  
• conserved.

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SHM and quadratic potentials

• SHM will occur whenever the potential is
  quadratic.
• Generally, this will not be the case
• For example, the potential betweenH atoms in an
  H2 molecule lookssomething like this

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SHM and quadratic potentials...

• However, if we do a Taylor expansion of this
  function about the minimum, we find that for
  smalldisplacements, the potential IS
  quadratic

U?(x0) 0 (since x0 is minimum of potential)
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SHM and quadratic potentials...
U(x) U?? (x0) x? 2 Let k U?? (x0)
Then U(x) k x? 2
U
U
x0
x
x ?
SHM potential!!
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Problem Vertical Spring

• A mass m 102 g is hung from a vertical spring.
  The equilibrium position is at y 0. The mass
  is then pulled down a distance d 10 cm from
  equilibrium and released at t 0. The measured
  period of oscillation is T 0.8 s.
• What is the spring constant k?
• Write down the equations for the position,
  velocity, and acceleration of the mass as
  functions of time.
• What is the maximum velocity?
• What is the maximum acceleration?

k
y
0
-d
m
t 0
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Problem Vertical Spring...

• What is k ?

k
y
0
-d
m
t 0
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Problem Vertical Spring...

• What are the equations of motion?
• At t 0,
• y -d -ymax
• v 0
• So we conclude

y(t) -d cos(?t) v(t) ?d sin(?t) a(t) ?2d
cos(?t)
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Problem Vertical Spring...
y(t) -d cos(?t) v(t) ?d sin(?t) a(t) ?2d
cos(?t)
?t
0
?
??
k
y
xmax d .1m vmax ?d (7.85 s-1)(.1m)
0.78 m/s amax ?2d (7.85 s-1)2(.1m) 6.2
m/s2
0
-d
m
t 0
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Transport Tunnel

• A straight tunnel is dug from Urbana through the
  center of the Earth and out the other side. A
  physics 111 student jumps into the hole at noon.
• What time does she get back to Urbana?

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Transport Tunnel...
where MR is the mass inside radius R
FG
R
RE
MR
but
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Transport Tunnel...
FG
R
RE
MR
Like a mass on a spring with
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Transport Tunnel...
Like a mass on a spring with
FG
R
RE
plug in g 9.81 m/s2 and RE 6.38 x 106 m get
? .00124 s-1 and so T 5067 s
84 min
MR
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Transport Tunnel...

• So she gets back to Urbana 84 minutes later, at
  124 p.m.

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Transport Tunnel...

• Strange but true The period of oscillation does
  not require that the tunnel be straight through
  the middle!! Any straight tunnel gives the same
  answer, as long as it is frictionless and the
  density of the Earth is constant.

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Transport Tunnel...

• Another strange but true fact An object orbiting
  the earth near the surface will have a period of
  the same length as that of the transport tunnel.

a ?2R 9.81 ?2 6.38(10)6 m ?
.00124 s-1 so T 5067 s
84 min
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Simple Harmonic Motion Summary
k
s
0
m
Force
Solution s A cos(?t ?)
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Recap of todays lecture

• Recap of last lecture
• Using initial conditions to solve problems
  (Text 14-1)
• The general physical pendulum (Text 14-3)
• The torsion pendulum
• Energy in SHM (Text 14-2)
• Atomic Vibrations
• Problem Vertical Spring (Text 14-3)
• Problem Transport Tunnel
• SHM Review
• Look at textbook problems Chapter 14 51, 53,
  57, 58, 65, 125