PHYS 20 LESSONS
Unit 6: Simple Harmonic
Motion
Mechanical Waves
Lesson 5: Pendulum Motion as SHM
Reading Segment #1:
Kinematics and Dynamics ofPendulum Motion
To prepare for this section, please read:
Unit 6: p. 13
Kinematics and Dynamics of Pendulum Motion
You will soon discover that pendulum motion can be equivalent to SHM.
However, before you can see this, you need to understand the nature of the forces and acceleration during pendulum motion.
Pendulum Motion Applet
To analyze the velocity, acceleration, and forces during pendulum motion, click on the following link:
Instructions:
Click on mass and move pendulum to an angle Click on “Play” and it will oscillate Click on Button 1 to see the forces Click on Button 2 to see the velocity and
acceleration
http://canu.ucalgary.ca/map/content/torque/aboutanaxis/apply/physicalpendulum/applet.html
Consider moving a pendulum to a deviation angle and then releasing it.
rest
FT
Fg = mg
rest
First, we draw the forces on the mass:
The force of gravity always acts downward.
Tension force acts along the string and towardsthe pivot (i.e. away from the mass).
FT
Fg = mg
rest
y
x
Next, we establish the y-axis along the string.
Notice that Fg is the diagonal force here.
FT
Fg = mg
rest
Fg y
Fg x
Next, we draw the x- and y-components of Fg.
FT
Fg = mg
rest
Fg y
Fg x
mg
Fxgsin
sinmgFxg
mg
Fyg
cos
cosmgFyg
We can determine each component as follows:
FT
Fg = mg
mg cos
mg sin
rest
The two “vertical” forces are equal but opposite in this position, and so there is no acceleration along the y-axis.
FT
Fg = mg
mg cos
mg sin
rest
a
The net force is mg sin , and so the pendulum’s acceleration is entirely along the x-axis.
That is, the acceleration is tangential.
FT
Fg = mg
mg cos
mg sin (Fs)
rest
a
mg sin acts in a direction opposite (approximately) to the displacement x, much like a spring force.
It is trying to return the mass to the equilibrium position (i.e. a restoring force).
x
FT
Fg = mg
mg cos
mg sin (Fs)
xrest
av
The pendulum speeds up until it reaches equilibrium position, where it has achieved maximum speed.
Equilibrium: Max speed
FT
v
Fg
Again, we draw the forces on the mass:
The force of gravity always acts downward.
Tension force acts along the string and towardsthe pivot (i.e. away from the mass).
FT
Fg = mg
mg cos
mg sin (Fs)
xrest
a
FT
v
Fg
At this moment, the mass is in uniform circular motion.
The acceleration is entirely vertical and towards the centre, and so it is centripetal.
ac
FT
Fg = mg
mg cos
mg sin (Fs)
xrest
a
FT
Fg = mg
mg cos
x rest
aac
FT
v
Fg
mg sin (Fs)
It is very similar when it moves back up.
mg sin acts to slow down the mass until it comes to rest at its maximum displacement.
FT
Fg = mg
mg cos
mg sin (Fs)
xrest
a
Reading Segment #2:
Pendulum Motion as SHM
To prepare for this section, please read:
Unit 6: pp. 13 - 14
Pendulum Motion as SHM
Consider a pendulum deviated at an angle .
x
Fs = mg sin
L
Again, we consider mg sin as being equivalent to the restoring force Fs.
x
Fs = mg sin
LL
xsin
For the triangle shown:
x
Fs = mg sin
L
L
xsin
Combining this with
sinmgFs
we discover that
L
xmgFs
or
xL
mgFs
x
Fs = mg sin
L
xL
mgFs
Since m, g, and L are constant for the pendulum, it follows that
xFs
Thus, the “spring force” has a direct relationship with the displacement x.
x
Fs = mg sin
L
However, technically speaking, for the pendulum to be in SHM, the restoring force must be directly proportional to the displacement along the arc.
This is not actually true, so a pendulum does not undergo SHM.
a
< 15
x a
Fs = mg sin
L
However, if the angle of deflection is less than 15, then we can safely assume that the displacement is equal to the arclength.
Under this condition, Fs x and the pendulum undergoes SHM.
a
Period of Pendulum Motion
As we have seen, for small amplitudes ( < 15), we can consider mg sin as a restoring force, much like Fs .
x
Fs = mg sin
L
As a result, since this is equivalent to a mass-spring system, we can use Hooke’s Law (Fs = k x) to create a formula.
x
Fs = mg sin
L
Starting with the formula we derived earlier:
xL
mgFs
We now substitute Hooke’s Law Fs = k x. This leaves us with
xL
mgxk
x
Fs = mg sin
L
Thus,
xL
mgxk
L
mgk
or,
mgLk
If we substitute this into the period formula for a mass-spring system, we discover the following:
k
mT 2
but,
g
L
k
mmgLk or
So,
g
LT 2
Equation:
The period T of a pendulum is calculated using the formula
SI Units: s
where
L is the length of the pendulum (in m)
g is the magnitude of the acceleration due to gravity (in m/s2)
g
LT 2
Note:
L is measured from the axis of rotation to the centre of the mass.
g
LT 2
For this formula to be used, < 15
L
axis
Example
A pendulum oscillates at a frequency of 0.42 Hz. Determine its length.
Try this example on your own first.Then, check out the solution.
= 2.38 s
Find period:
Tf
1
fT
1
42.0
1
Find length:
g
LT 2
g
LT
2
2
2
g
LT
2
g
LT
2
22
2
g
LT
g
LT
2
2
4
g
LT
2
2
4
LTg 22 4
2
2
4Tg
L
2
2
4Tg
L
2
2
4
38.281.9
= 1.4 m
Practice Problems
Try these problems in the Physics 20 Workbook:
Unit 6 p. 66 #1 - 4
Reading Segment #3:
Energetics of Pendulum Motion
To prepare for this section, please read:
Unit 6: p. 15
Energetics of Pendulum Motion
Again, consider moving a pendulum to a deviation angle and then releasing it.
rest
Pendulum Motion Applet
Analysis of Pendulum Motion
To analyze the energy of pendulum motion, click on the following link:
Note: Click on “show” if you wish to see theacceleration vector and its components.
rest
= A
At the position of maximum displacement, the angle of deviation represents the amplitude of oscillation.
i.e. max = A
rest
A
hmax
Equilibriumposition
At maximum displacement, the mass is at a maximum height (relative to the equilibrium position).
Thus, it has maximum gravitational potential energy.
Max Epg
Ref h: h = 0
rest
A
hmax
Equilibriumposition
The mass is also at rest, and so it has no kinetic energy.
Max Epg
Ek = 0
rest
A
hmax
Equilibriumposition
The total mechanical energy at this position is
Max Epg
Ek = 0
kpm EEEg
(max)gpE The mass only has gravitational potential energy
rest
vmax
A
hmax
Equilibriumposition
Max Epg
Ek = 0
As the pendulum falls, it speeds up.
When it reaches the equilibrium position, the mass has attained its maximum speed.
rest
vmax
A
hmax
Equilibriumposition
Max Epg
Ek = 0
Since the speed is at a maximum, the mass has maximum kinetic energy.
Max Ek
rest
vmax
A
hmax
Equilibriumposition
Max Epg
Ek = 0
However, since the mass has reached the equilibrium position, it has no gravitational potential energy.
Ref h: h = 0
Max Ek
Epg = 0
rest
vmax
A
hmax
Equilibriumposition
Max Epg
Ek = 0
Max Ek
Epg = 0
The total mechanical energy at the equilibrium position is
kpm EEEg
(max)kE The mass only has kinetic energy
rest
vmax
A
hmax
A
Equilibriumposition
Max Epg
Ek = 0
Max Ek
Epg = 0
Max Epg
Ek = 0rest
It is similar when it moves up to maximum displacement.
The mass slows down until it comes to rest at its maximum height. At this position, it has no kinetic energy and maximum gravitational potential energy.
Maximum displacement
Energy Formulas of Pendulum Motion
To derive an energy formula for pendulum motion, we compare maximum displacement with equilibrium position.
rest
vmax
A
hmax
Equilibriumposition
Max Epg
Ek = 0
Max Ek
Epg = 0
If this is an ideal pendulum, then there is no energy lost due to friction or air resistance. rest
vmax
A
hmax
Equilibriumposition
Max Epg
Ek = 0
Max Ek
Epg = 0Thus, the total mechanical energy must remain constant.
i.e.
Em (top) = Em (bottom)
rest
vmax
A
hmax
Equilibriumposition
Max Epg
Ek = 0
Max Ek
Epg = 0
Em (top) = Em (bottom)
max(max) kp EEg
rest
vmax
A
hmax
Equilibriumposition
Max Epg
Ek = 0
Max Ek
Epg = 0
Em (top) = Em (bottom)
max(max) kp EEg
2maxmax 2
1mvmgh
rest
vmax
A
hmax
Equilibriumposition
Max Epg
Ek = 0
Max Ek
Epg = 0
Em (top) = Em (bottom)
max(max) kp EEg
2maxmax 2
1mvmgh
2maxmax 2
1vgh
This is not on the formula sheet, and so you must derive this each time.
Example
On a different planet, an 80 cm long pendulum oscillates with a period of 2.9 seconds.
If this pendulum reaches a maximum height of 1.5 cm (above its equilibrium position), then determine its maximum speed.
Assume no air resistance or friction.
Try this example on your own first.Then, check out the solution.
Find the acceleration due to gravity:
g
LT 2
g
LT
2
2
2
g
LT
2
g
LT
2
22
2
g
LT
g
LT
2
2
4
g
LT
2
2
4
LTg 22 4
2
24
T
Lg
2
24
T
Lg
2
2
9.2
80.04
= 3.755 m/s2
Find the maximum speed:
Em (top) = Em (bottom)
max(max) kp EEg
rest
vmax
A
hmax
Equilibriumposition
Max Epg
Ek = 0
Max Ek
Epg = 0
Em (top) = Em (bottom)
max(max) kp EEg
2maxmax 2
1mvmgh
Em (top) = Em (bottom)
max(max) kp EEg
2maxmax 2
1mvmgh
2maxmax 2
1vgh
2maxmax 2
1vgh
2maxmax2 vgh
2maxmax 2
1vgh
2maxmax2 vgh
maxmax 2ghv
maxmax 2ghv
015.0755.32
= 0.34 m/s
Practice Problems
Try these problems in the Physics 20 Workbook:
Unit 6 p. 6 #11