Transcript Potential Energy - McMaster University

Simple Harmonic
Motion (V)
Circular Motion
The Simple Pendulum
Physics 1D03 - Lecture 34
1
SHM and Circular Motion
Uniform circular motion about in the xy
plane, radius A, angular velocity  :
(t) = 0 +  t
(similar to, x=xo+vt)
A

and so
x  A cos  A cos( 0  t )
y  A sin   A sin( 0  t )
Hence, a particle moving in one dimension can be expressed
as an ‘imaginary’ particle moving in 2D (circle), or vice versa the ‘projection’ of circular motion can be viewed as 1D motion.
Physics 1D03 - Lecture 34
2
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.
Physics 1D03 - Lecture 34
3
Phase Constant, θo
For circular motion,
the phase constant
is just the angle at
which the motion started.
A

o
Physics 1D03 - Lecture 34
4
Example
An object is moving in circular motion with an angular
frequency of 3π rad/s, and starts with an initial angle of
π/6. If the amplitude is 2.0m, what is the objects angular
position at t=3sec ?
What are the x and y values of the position at this time ?
Physics 1D03 - Lecture 34
5
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
6
Simple Pendulum
Using sin(θ)~θ for small angles, we have the
following equation of motion:
L
d 2
g
 
2
dt
L
Which give us:
θ
T

g
L
------------------------------------------------------------------------Hence:
or:
gT 2
L 2 

4 2
g
mg
2
4

L
2
g  L 
T2
Application - measuring height
- finding variations in g → underground resources
Physics 1D03 - Lecture 34
7
Actually:
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
d 2
2




2
dt
Physics 1D03 - Lecture 34
8
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
(2 / period)
Physics 1D03 - Lecture 34
9
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?)
Physics 1D03 - Lecture 34
10
Application
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?
?
Physics 1D03 - Lecture 34
11
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
12
SHM and Damping – EXTRA !!!
SHM: x(t) = A cos ωt
Motion continues indefinitely.
Only conservative forces act,
so the mechanical energy is
constant.
Damped oscillator: dissipative
forces (friction, air resistance, etc.)
remove energy from the oscillator,
and the amplitude decreases with
time.
x
t
x
t
Physics 1D03 - Lecture 34
13
A damped oscillator has external nonconservative
force(s) acting on the system. A common example
is a force that is proportional to the velocity.
f = bv where b is a constant damping coefficient
F=ma give:
dx
d 2x
 kx  b  m 2
dt
dt
For weak damping (small b), the solution is:
x
x(t )  Ae

b
t
2m
eg: green water
cos(t   )
A e-(b/2m)t
t
Physics 1D03 - Lecture 34
14