Harmonic Motion (III)
•Simple and Physical Pendulum
•SHM and uniform circular motion
Physics 1D03 - Lecture 34
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Simple Pendulum
Gravity is the “restoring force” taking the place of the
“spring” in our block/spring system.
L
Instead of x, measure the displacement as the arc length
s along the circular path.
θ
T
Write down the tangential component of F=ma:
Restoring force  mg sin 
d 2s
m 2  m at  m g sin( )
dt
But s  L
mg
s
mg sin θ
d 2
g
 2   sin 
dt
L
Physics 1D03 - Lecture 34
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SHM:
Simple pendulum:
d 2x
2



x
2
dt
d 2
g
  sin 
2
dt
L
The pendulum is not a simple harmonic oscillator!
However, take small oscillations:
sin    (radians) if  is small.
Then
d 2
g
g
  sin    
2
dt
L
L
Physics 1D03 - Lecture 34
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For small  :
This looks like
d 2
g
 
2
dt
L
d 2x
2



x , with angle  instead of x.
2
dt
The pendulum oscillates in SHM with an angular
frequency

g
L
and the position is given by  (t )   o cos(t   )
phase
constant
amplitude
(2p / period)
Physics 1D03 - Lecture 34
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• a simple harmonic oscillator is a mathematical
‘approximation’ to the full problem
• for large amplitudes, the solution that the SHO gives
us start to deviate from what they actually should be :

l  1 2  o 
T  2p
1  sin    

g 4
2

Unlike the SHO, the actual solution depends on the amplitude!
Physics 1D03 - Lecture 34
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Quiz
Pendulum clocks (“grandfather clocks”) often have a
swinging arm with an adjustable weight. Suppose
the arm oscillates with T=1.05sec and you want to
adjust it to 1.00sec. Which way do you move the
weight?
A) Up
B) Down
?
Physics 1D03 - Lecture 34
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Question: A simple pendulum hangs from the ceiling
of an elevator. If the elevator accelerates
upwards, the period of the pendulum:
a) Gets shorter
b) Gets larger
c) Stays the same
Question: What happens to the period of a simple
pendulum if the mass m is doubled?
Physics 1D03 - Lecture 34
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Question: A geologist is camped on top of a large deposit
of nickel ore, in a location where the gravitational field is
0.01% stronger than normal. the period of his pendulum
will be
a) longer
b) shorter
(and by how much, in percent?)
Question: How high is the ceiling?
Physics 1D03 - Lecture 34
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Simple pendulum: A particle
on a massless string.

“Physical” pendulum: any
rigid body, pivoted at P, and free
to swing back and forth.
P
To find the period:
1) Consider the torque due to gravity

CM
2) Write t()  Ia  I (d2/dt2)
3) SHM if t is proportional to 
mg
Physics 1D03 - Lecture 34
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Example: a metre stick, pivoted at one end. What is
its period of oscillation?
“Uniform thin rod, pivot at end”: I = 1/3
ML2
Calculate torque about the end:

t  Ia
L
L 1
 Mg sin    ML2  a
2 3
Mg
and so
d 2
3g
a  2 
sin 
dt
2L
Note, this does not describe SHM!
Physics 1D03 - Lecture 34
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But, for small oscillations, sin   
so
d 2
3g
2







2
dt
2L
This looks like
d 2x
2



x
2
dt
The angular frequency is
and the period is
, with angle  instead of x.

3g
2L
2p
2L
T
 2p

3g
This is like a simple pendulum of length 2/3L.
Physics 1D03 - Lecture 34
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Any swinging object can be analysed in a similar way; we just need
to know its moment of inertia, I, about the pivot point.
d 2
 Mgd sin   I 2
dt
Mgd

I
d
θ
Mg
in general
For a simple pendulum, I = Ml2
and  
g
l
 (t )   o cos(t   )
phase
constant
amplitude
(2p / period)
Physics 1D03 - Lecture 34
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Quiz: What happens to the period of the
metre stick when the pivot is moved closer
to the centre?
A)
B)
C)
D)
The period gets longer.
The period gets shorter.
The period stays the same.
It’s rather difficult to tell.
P
mg
Physics 1D03 - Lecture 34
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
2p
Mgd
 2p
, or T 

I
I
Mgd
I is the moment of inertia about the pivot.
From the parallel-axis theorem:
I  I CM  Md 2
 121 ML2  Md 2
So
T  2p
L2 d

gd
g
1
12
(for a uniform thin rod).
(for a uniform thin rod).
Physics 1D03 - Lecture 34
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SHM and Circular Motion
Uniform circular motion about in the
xy plane, radius A, speed v, and
angular velocity  = v/A :
(t) = 0 +  t
A

and so
x  A cos  A cos( 0  t )
y  A sin   A sin( 0  t )
Real particle moves on the x axis
“Imaginary” particle moves in a circle.
Physics 1D03 - Lecture 34
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x  A cos  A cos(o  t )
y  A sin   A sin(o  t )
Compare with our expression for 1-D SHM.
x  A cos(t  )
Result:
SHM is the 1-D projection
of uniform circular motion.
An interesting side problem: (try this out on your own)
Start with

r  A cosi  A sin j
Differentiate twice, and then show
 v2
ar 
r
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Summary
The projection of uniform circular motion onto an axis is
SHM in 1-D.
The oscillation of a simple pendulum is approximately
SHM, if the amplitude is small, with angular frequency

g
L
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Practice problems, Chapter 15
3, 5, 11, 19, 23, 31, 67
(6th ed – Chapter 13)
1, 3, 5, 9, 19, 23, 29, 67
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