Harmonic Motion (III)

1
Harmonic Motion (III)

• Simple and Physical Pendulum
• SHM and uniform circular motion

2
Simple Pendulum
Gravity is the restoring force taking the place
of the spring in our block/spring
system. Instead of x, measure the displacement
as the arc length s along the circular
path. Write down the tangential component of
Fma
L ? T
Restoring force
mg
s
mg sin ?
3
SHM
Simple pendulum
The pendulum is not a simple harmonic oscillator!
4
For small q
, with angle q instead of x.
This looks like
The pendulum oscillates in SHM with an angular
frequency
and the position is given by
phase constant
amplitude
(2p / period)
5

• 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

Unlike the SHO, the actual solution depends on
the amplitude!
6
Quiz
Pendulum clocks (grandfather clocks) often have
a swinging arm with an adjustable weight.
Suppose the arm oscillates with T1.05sec and you
want to adjust it to 1.00sec. Which way do you
move the weight?
A) UpB) Down
?
7
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?
8
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
• shorter

(and by how much, in percent?)
Question How high is the ceiling?
9
Simple pendulum A particle on a massless string.
q
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
• 2) Write t(q) Ia I (d2q/dt2)
• 3) SHM if t is proportional to -q

q
CM
mg
10
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
L
q
and so
Mg
Note, this does not describe SHM!
11
But, for small oscillations, sin q ? q
so
, with angle q instead of x.
This looks like
The angular frequency is
and the period is
This is like a simple pendulum of length 2/3L.
12
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 ? Mg
in general
For a simple pendulum, I Ml2

and
phase constant
amplitude
(2p / period)
13
Quiz What happens to the period of the metre
stick when the pivot is moved closer to the
centre?

1. The period gets longer.
2. The period gets shorter.
3. The period stays the same.
4. Its rather difficult to tell.

14
, or
I is the moment of inertia about the pivot.
From the parallel-axis theorem
(for a uniform thin rod).
So
(for a uniform thin rod).
15
SHM and Circular Motion
Uniform circular motion about in the xy plane,
radius A, speed v, and angular velocity w v/A
q(t) q0 w t and so
A
q
Real particle moves on the x axis Imaginary
particle moves in a circle.
16
Compare with our expression for 1-D SHM.
Result An interesting side problem (try
this out on your own) Start with Differentiate
twice, and then show
SHM is the 1-D projection of uniform circular
motion.
17
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
18
Practice problems, Chapter 153, 5, 11, 19, 23,
31, 67(6th ed Chapter 13) 1, 3, 5, 9, 19, 23,
29, 67