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