Episode 305: Energy in SHM
Qualitatively, students will appreciate that there is a continuous interchange between potential
and kinetic energy during SHM. Here, they can also learn about the mathematical basis for
calculating energy.
Summary
Demonstration: An experimental displacement-time graph. (10 minutes)
Discussion: Maximum values of quantities in SHM. (15 minutes)
Student questions: Practice with the equations. (30 minutes)
Discussion: Energy changes in SHM. (20 minutes)
Student questions: Energy of a pendulum. (20 minutes)
Demonstration:
An experimental displacement-time graph
Use a ‘water pendulum’ to draw a large displacement-time graph for a pendulum. You could ask a
group of students to prepare this in advance and demonstrate it to the class.
TAP 305-1: The water pendulum
Discussion:
Maximum values of quantities in SHM
Refer back to the sin and cos equations for SHM. Show that the maximum values of
displacement, velocity and acceleration are given by (the term in front of sin or cos):

maximum displacement = A

maximum velocity = A

maximum acceleration = A2

Compare these relationships with the equations for circular motion:

displacement = r

velocity = r

acceleration = r2
If you have adopted the ‘auxiliary circle’ approach earlier, the parallels should be clear.
1
Student questions:
Practice with the equations
It will help to provide some more practice in using the equations and analyzing motion.
TAP 305-2: Oscillators
Discussion:
Energy changes in SHM
Think about the energy changes in a mechanical oscillator. Recap that, as it passes through its
equilibrium position, its speed and hence its kinetic energy are a maximum. At the maximum
displacement, the speed and hence the kinetic energy are both zero. The potential energy will be
a maximum when the speed is zero and vice versa. Assuming that there is no friction or air drag
the total energy input E of the oscillator must remain constant
For the mass and spring system, the work done stretching a spring by an amount x is the area
under the force extension graph = 1/2 kx2. The
total energy
PE-extension graph is a parabola.
The kinetic energy will be zero at +A and a
maximum when x = 0, so its graph is an inverted
version of the strain energy graph. At any
position kinetic + elastic strain energy is a
constant E, where E = KEmax = PEmax.
kinetic energy
PEmax  A2, so the total energy E of SHM is
proportional to (amplitude) 2.
potential energy
TAP 305-3: Elastic energy
-A
0
Draw a graph from x = {–A to +A}, to show kinetic
energy, strain energy and total energy. You can also draw graphs of KE and PE against time.
TAP 305-4: Energy flow in an oscillator
Student questions:
Energy of a pendulum
Some useful questions on energy of a pendulum.
TAP 305-5: Energy and pendulums
2
+A
TAP 305- 1: The water pendulum
As inky water runs out of a hole in the bottom of a can, it can create a trace on paper slowly
moved underneath. Although this is a ‘rough and ready’ way of recording the oscillation, it can
give you useful information.
You will need

water pendulum

four sheets of graph paper taped together, or light-coloured sugar paper cut to size

two retort stands, bosses and clamps

two G clamps, approximately 5 cm jaw

ink

hand-held stopwatch
string
~1.5 m
pull graph
paper at right
angles to
movement of
pendulum
What to do
You want the pendulum to swing with an amplitude of half the width of the graph paper. As it is
swinging, you are going to drag the paper underneath the dribbling water to create a time trace.
To do this you will need to pull the paper with constant velocity, so do a few ‘dry runs’ before
putting water in the can.
Make sure the end of the tube is as near as possible to the paper at the bottom of the swing.
1.
Put your finger over the hole and put about 2 cm of water in the can. Now add 10 drops
of ink and allow it to mix.
3
2.
Set the pendulum swinging and slowly pull the graph paper under the water. This will give
you a time trace. Let it dry and add axes to the line or make a good copy on another
sheet of graph paper.
Now you have done the experiment, think how to answer these questions:

How can you mark time values on the graph?

Does the periodic time of the oscillation change as the water runs out?

What is the maximum velocity of the pendulum?
You have seen
1.
The pendulum reaches its maximum velocity when it travels through its mean position.
2.
The time trace is a displacement–time graph so velocity at any instant can be found by
taking the gradient at that instant.
Practical Advice
If a little care is taken, this activity will yield a reasonable time trace. The aim of this activity is
getting your students to think more carefully about the nature of oscillators. Producing a time
trace should encourage students to observe more carefully. The graph obtained should be clear
enough for them to consider how the gradient varies over the cycle.
The ‘water pendulum’ needs to be made up. At its simplest it is just a can with a 1 mm hole in the
centre. A more reliable effect is produced when a glass tube is inserted into a bung and then the
bung in the bottom of the can.
The supporting strings should be about 1.5 m long.
Safety
If the paper is on the bench, the string supports are roughly 2.5 m from the floor. This requires
two persons; one to hold the ladder or steps and the other to fix the string supports. If the paper is
on the floor, the retort stands can be on two tables; ensure that they cannot move.
External reference
This activity is taken from Advancing Physics chapter 10, 200P
4
TAP 305- 2: Oscillators
Questions
Here is the displacement–time graph of an oscillator.
0
A B
0
C
D
E
F
time
1.
Consider the speed of the oscillator at the four times labelled A, B, C, and D. Arrange the
times A, B, C, D in order of decreasing speed.
2.
How does the velocity at time B compare with that at time E?
3.
How does the velocity at time D compare with that at time F?
4.
At which of the times 0 to F is the acceleration at its largest value?
5.
At which of the times 0 to F is the displacement equal in size to the amplitude of the
motion?
5
6.
Consider the time intervals 0–B, 0–D, 0–F, B–E, D–F. If the periodic time of the oscillator
is T, write down each interval in terms of T. (0–F = 3 T is the sort of answer expected,
though this particular answer would be wrong.)
Here are three things which would oscillate in a laboratory on Earth.
(a)
(b)
(c)
piston
air
7.
Which, if any, would oscillate in a spacecraft going at steady speed a long way from the
Earth and from any planet or star?
Explain your answer.
Sketch a large displacement–time graph for two periods of a simple harmonic motion.
8.
Mark with M any instant where the speed is a maximum
6
9.
Mark with Z any instant when the speed is zero.
10.
Mark places where the acceleration is high H and where it is low L.
Answer each of the following, giving reasons:
11.
Does a tuning fork, used by musicians, vibrate with simple harmonic motion?
12.
Is the bouncing of a ball a simple harmonic motion?
13.
If a pendulum were taken to the top of a mountain, would it gain or lose time?
7
Practical advice
These questions are concerned with basic ideas about simple harmonic motion. They may
therefore prove useful for class discussion. Others provide practice identifying and using relevant
equations.
Answers
1.
BCAD
2.
Same magnitude, opposite direction.
3.
Both zero.4.
5.
0, D and F
6.
0–B = T 4, 0–D = T/2, 0–F = T, B–E =T/2, D–F = T/2
7.
(b) will oscillate. Only the horizontal mass-spring system does not depend on gravity.
0, D and F
8., 9. and 10.
Note again that a is proportional to displacement, but in the opposite direction.
Z,H
Z,H
L
M
L
L
M
L
M
L
L
M
L
Z,H
M
L
Z,H
t
11.
Yes. For small displacements, this is an instance of Hooke’s law, with the force
proportional to the displacement.
12.
No. The force (and acceleration) is constant when removed from the surface, being the
weight of the ball.
13.
Loses time.
T  2
l
g
and g decreases with height (distance from Earth’s centre), so T will increase.
External reference
This activity is taken from Advancing Physics chapter 10, 160S
8
TAP 305- 3: Elastic energy
The relationship between force to extend a spring, and extension, determines the energy stored.
Energy stored in a stretched spring
area below graph
= sum of force 
change in displacement
extra area
F1 x
F1
total area
1 Fx
2
0
x
0
unstretched
extension x
force F1
work F1 x
no force
larger force
Energy supplied
small change x
energy supplied = F x
F=0
x=0
F = kx
x
stretched to extension x by force F:
energy supplied = 12 Fx
spring obeys
Hooke’s law: F = kx
energy stored in stretched
spring = 12 kx2
Energy stored in stretched spring is 12 kx2
9
Practical advice
This diagram is reproduced here so that you can talk through it, or adapt it to your own purposes.
External reference
This activity is taken from Advancing Physics chapter 10, 120O
10
TAP 305- 4: Energy flow in an oscillator
The energy sloshes back and forth between being stored in a spring, and carried by the mass.
Energy flow in an oscillator
displacement
potential energy
= 12 ks2
0
s = A sin 2ft
time
energy in stretched spring
potential energy
0
PE =
1
2
kA2 sin22ft
time
mass and
vmax
spring
oscillate
A
vmax
A
vmax
energy carried by moving mass
kinetic energy
0
KE =
1
2
2
mvmax
cos22ft
time
velocity
kinetic energy
= 12 mv2
v = vmax cos 2ft
0
vmax = 2fA
time
from spring to
moving mass
energy in
stretched spring
from spring to
moving mass
energy in
moving mass
from moving
mass to spring
from moving
mass to spring
The energy stored in an oscillator goes back and forth between stretched spring and moving
mass, between potential and kinetic energy
11
Practical advice
This diagram is reproduced here so that you can talk through it, or adapt it to your own purposes.
External reference
This activity is taken from Advancing Physics chapter 10, 130O
12
TAP 305- 5: Energy and pendulums
Questions
A body of mass 100 g undergoes simple harmonic motion with amplitude of 20 mm. The
maximum force which acts upon it is 0.05 N. Calculate:
1.
Its maximum acceleration.
2.
Its period of oscillation.
A baby in a ‘baby bouncer’ is a real-life example of a mass-on-spring oscillator. The baby sits in a
sling suspended from a stout rubber cord, and can bounce himself up and down if his feet are just
in contact with the ground. Suppose a baby of mass 5.0 kg is suspended from a cord with spring
–1
constant 500 N m–1. Assume g = 10 N kg .
3.
Calculate the initial (equilibrium) extension of the cord.
4.
What is the value of  (= 2/T)?
5.
The baby is pulled down a further distance, 0.10 m, and released. How long after his
release does he pass through his equilibrium position?
6.
What is the maximum speed of the baby?
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A simple pendulum has a period of 4.2 s. When it is shortened by 1.0 m the period is only 3.7 s.
7.
Without assuming a value for g, calculate the original length of the pendulum.
8.
Calculate the acceleration due to gravity g suggested by the data.
Practical advice
These are rather harder questions on simple harmonic motion, involving numerical, and in the
case of the last two, algebraic competence.
Answers and worked solutions
1.
a
2.
0.05 N
F

 0.5 m s 2
m
10 kg
for 10 kg read 0.1 kg
a  2 s
and

2
T
so
T  2
20  10 3 m
s
 2
 1.3 s.
a
0 .5 m s  2
3.
x
F mg 5.0 kg  10 N kg –1


 0. 1 m
x
x
500 N m –1

2
T
4.
and
14
T  2
m
k
so

500 N m 1
 10 rad s 1
5.0 kg
k

m
5.
T 
2
2


10 rad s 1
and
T
2

 0.16 s
4 4  10 rad s 1
6.
v max  2fA  A  10 rad s 1  0.10 m  1.0 m s 1 .
7.
With two unknowns, this requires the use of simultaneous equations. Initially
l
T  2
 4.2 s
g
When the length is shorter
2
l  1.0 m
 3.7 s.
g
Dividing one equation by the other
l
g
l  1 .0 m
g

4 .2 s
.
3 .7 s
This gives
2
l
 4.2 

  1.29
l  1.0 m  3.7 
l  1.29(l  1.0 m)  1.29l  1.29 m
and
0.29l  1.29 m
so
l 
1.29 m
 4.5 m.
0.29
15
8.
Since
l
g
T  2
g 
4 2 l
T
2

4 2 4.5 m
4.2 s
2
 10 m s 2 .
External reference
This activity is taken from Advancing Physics chapter 10, 170S
16