1. ELASTICITY AND
HARMONIC MOTION
Standard Competency
Analyzes the nature phenomenon and its regularity within the
scope of particle’s Mechanics
Base Competency
Analyzes the effect of force on the elasticity properties of a
material
Learning Objectives
1 Describes the characteristic of force on elastic material
base on experiment performed
2 Identifies the elastic moduli and spring force constant
3 Compares the force constant base on observatiob’s data
4 Analyses the series and parallel spring configuration
5 Calculates spring’s elongation
6 Determines the value of spring’s force constant
References
[1] John D Cutnell dan Kenneth W. Johnson (2002). Physics 5th Ed
with Compliments. John Wiley and Sons, Inc. hal. 273-294
[2] Sunardi dan Etsa Indra Irawan (2007). Fisika Bilingual SMA/MA
untuk SMA/MA Kelas XI. CV Yrama Widya hal. 255-298

2. ELASTICITY
Elasticity is the property or ability of an object or material to
restored to its original shape after apllied distortion vanished.
Elasticity ≠ Plastics
A spring is an example of an elastic object - when stretched, it
exerts a restoring force which tends to bring it back to its original
length. This restoring force is generally proportional to the
amount of stretch, as described by Hooke's Law. For wires or
columns, the elasticity is generally described in terms of the
amount of deformation (strain) resulting from a given stress
(Young's modulus). Bulk elastic properties of materials describe
the response of the materials to changes in pressure.

3. Elastic Moduli
When a force is exerted on the
suspended metal, the length of the
object changes. As long as the
amount of elongation, ΔL, is small
compared to its length, the elongation
is directly proportional to the force.
This was first noted by Robert Hooke.
Hooke’s Law of Elasticiy
“Within object’s elasaticity limit, the applied force F is
proportional to the object elongation ΔL ”
F = −k ΔL
(−) minus sign shows that the restoring force F is oppose to
object’s direction
STRESS (TENSION)
Stress or tension is defined as force per unit cross section area.
It has unit (SI) N/m2.
F
σ =
A
STRAIN (SCRETCH)
Stress or tension is defined as ratio between elongation and
initial length. It is unitless.
Δl
e=
lo

4. YOUNG’s MODULUS
Young’s modulus or elastic modulus is ration between stress and
strain. It has unit (SI) N/m2.
σ 1 F
E = → ΔL = L
e E A
E is the elastic modulus or Young's modulus and is only
dependent on the material.
Physics Charts – Elastic, Shear and Bulk Moduli
Material Elastic Modulus Shear Modulus Bulk Modulus
E ( N/m2 ) G ( N/m2 ) B ( N/m2 )
Solids
Steel 200 x 10 9 80 x 10 9 140 x 10 9
Brass 100 x 10 9 35 x 10 9 80 x 10 9
Aluminum 70 x 10 9 25 x 10 9 70 x 10 9
Concrete 20 x 10 9
Brick 14 x 10 9
Bone (limb) 15 x 10 9 80 x 10 9
Liquids
Water 2.0 x 10 9
Alcohol 1.0 x 10 9
(ethyl)
Mercury 2.5 x 10 9
Gases
Air, He, H2, 1.01 x 10 5
CO2

5. How force is affecting a material in term of its elongation
described in graph below.
In the Plastic Region, the
material does not change
in a linear fashion.
If stretched to the Elastic
Limit or beyond, it does
not return to its original
length.
If stretched to the
Breaking Point, the
material will break into two
pieces.
Terms related to applied force on a material are tension,
compression and shear

6. Example
A metal (steel) rod whose has cross section area of 4 mm2 and
length of 40 cm is hang and pulled down by force of 100 N. If
the elastic modulus of metal is 2 x 1011 N/m2, calculate
(a) stress
(b) strain
(c) elongation length
Known
A = 4 mm2 = 4 x 106 m2
lo = 40 cm = 0.4 m
F = 100 N
E = 2 x 104 N/m2
Asked
(a) stress, σ
(b) strain, e
(c) elongation length, Δl
Answer
F 100
(a) σ = = −6
= 2.5 x 10− 6 N/m2
A 4 x 10
σ 2.5 x 107
(b) E = = 11
= 1.25 x 10− 4
e 2 x 10
(c) Δl = e x lo
= (1.25 x 10−4) (4 x 104)
= 5 x 105 m

7. EXERCISES
[1] A 15 cm long animal tendon was found to stretch 3.7 mm by
a force of 13.4 N. The tendon was approximately round with
an average diameter of 8.5 mm. Calculate the elastic
modulus of this tendon.
[2] How much pressure is needed to compress the volume of an
iron block by 0.10 percent? Express answer in N/m2, and
compare it to atmospheric pressure (1.0 x 105 N/m2).
[3] A depths of 2.00 x 10 3 m in the sea, the pressure is about
200 times atmospheric pressure. By what percentage does
an iron bathysphere's volume change at this depth?
[4] A nylon tennis string on a racquet is under a tension of 250.
N. If its diameter is 1.00 mm, by how much is it lengthened
from its untensioned length of 30.0 cm
[5] A vertical steel girder with a cross-sectional area of 0.15 m2
has a 1550 kg sign hanging from its end. (a) What is the
stress within the girder? (b) What is the strain on the
girder? (c) If the girder is 9.50 m long, how much is it
lengthened? (Ignore the mass of the girder itself.)
[6] A scallop forces open its shell with an elastic material called
abductin, whose elastic modulus is 2.0 x 10 6 N/m2. If this
piece of abductin is 3.0 mm thick, and has a cross-sectional
area of 0.50 cm2, how much potential energy does it store
when compressed 1.0 mm?

8. Answers
[1]
1 F
ΔL = Lo
E A
Convert Units and solve for E. F should be in N and area A
in m2. L and ΔL have same unit.
Lo = 15 cm ΔL = 0.37 cm r = 0.00425 m
A = r = ( 0.00425 m ) = 5.7 x 10 -5 m2
2 2
1 F 1 13.4 N
ΔL = Lo = 15 cm = 9.5 . 106 N/m2
E A 0.37 cm 5.7 . 10 m
−5 2
ΔV 1
[2] = − ΔP solve for ΔP
Vo B
ΔV − 0.10
ΔP = −B = − 90 . 109 N/m2 = 9.0 . 107 N/m2
Vo 100
9.0 . 107 N/m2
= 900
1.0 . 105 N/m2
It is 900 times greater than atmospheric pressure
ΔV 1 1
[3] = − ΔP = − 2.0 . 107 N/m2 = − 2.2 . 10 − 4
Vo B 9
90 . 10 N/m2
Percentage = - 2.2 x 10 -4 x 100 = - 2.2 x 10 -2

9. [4] A = πr2 = 3.14 (0.0005 m)2 = 7.9 x 10 -7 m2
1 F 1 250 N
ΔL = Lo = 30.0 cm = 1.9 cm
E A 5 . 10 Nm 7.9 . 10- 7 m2
9 -2
[5] F = mg = 1550 kg x 9.80 m s -2 = 15200 N
F 15200
stress = = = 1.0 . 105 N/m2
A 0.15
ΔL 1 1
strain = = stress = 1.0 . 105 N/m2 = 5.0 . 10 − 7
Lo E 9
200 . 10 N/m2
ΔL = Strain x Lo = 5.0 x 10 -7 x 9.50 m = 4.8 x 10 -6 m
1 F F
[6] ΔL = Lo and F = kΔL ⇒ ΔL =
E A k
PE = 1
2 k ΔL2 set ΔL = ΔL
1 F F
Lo = solved for k ; 1 m = 100 cm, 1 m2 = 1000 cm2
E A k
E A 2.0 . 10 6 N/m2 x 5.0 . 10-5 m2
k = = = 3.3 . 10 4 N/m
Lo -3
3.0 . 10 m
PE = ½ k ΔL2 = ½ (3.3 . 104 N/m)(1.0 . 10−3 m)2 = 0.017 J

10. ELASTICITY AND
HARMONIC MOTION
Standard Competency
Analyzes the nature phenomenon and its regularity within the
scope of particle’s Mechanics
Base Competency
Analyzes the relation between force and harmonic motion
Learning Objectives
1 Describes the characteristic of motion on vibrate spring
2 Explains the relation between the period of harmonic
motion and mass weighted base on observation’s data
3 Analyzes the displacement, velocity and acceleration
planetary motion within a universe base on Keppler’s Law
4 Analyses the potentian and mximum kinetic energy on
harmonic motion
References
[1] John D Cutnell dan Kenneth W. Johnson (2002). Physics 5th Ed
with Compliments. John Wiley and Sons, Inc. hal. 273-294
[2] Sunardi dan Etsa Indra Irawan (2007). Fisika Bilingual SMA/MA
untuk SMA/MA Kelas XI. CV Yrama Widya hal. 255-298

11. VIBRATION – SIMPLE HARMONIC MOTION (SHM)
Each day we encounter many kinds of oscillatory motion, such as
swinging pendulum of a clock, a person bouncing on a
trampoline, a vibrating guitar string, and a mass on a spring.
They have common properties:
1. The particle oscillates back and forth about a equilibrium
position. The time necessary for one complete cycle (a
complete repetition of the motion) is called the period, T.
2. No matter what the direction of the displacement, the force
always acts in a direction to restore the system to its
equilibrium position. Such a force is called a “restoring force”
In nearly all cases, at least for small displacement, there is an
“effective” restoring force that pulls back towards the
equilibrium position, proportional to the displacement.
Look at a mass on a spring as example
x = displacement
F = force due to spring
F = −k x
The restoring force is opposite to the displacement.

12. Give m a positive displacement where x = A. then release it. F
will pull the mass back towards x = 0.
The mass’s inertia will even change it back to x = −A.
Now the restoring force will be to
the right, where x is negative.
F pushes m back through x =
0, then the whole sequence
keeps repeating. This is refer as
vibration of simple harmonic
motion (SHM).
3. The number of cycles per
unit time is called the
“frequency” f.
1
f =
T
Unit: period (s)
frequency (Hz, SI unit),
1 Hz = 1 cycle/s
4. The magnitude of the
maximum displacement from
equilibrium is called the
amplitude, A, of the motion.

13. Simple harmonic motion (SHM):
An oscillating system which can be described in terms of sine
and cosine functions is called a “simple harmonic oscillator” and
its motion is called “simple harmonic motion”.
Equation of motion of the simple harmonic oscillator
Figure in the right shows
a simple harmonic
oscillator, consisting of a
spring of force constant K
acting on a body of mass
m that slides on a
frictionless horizontal
surface. The body moves
in x direction.
where k is the spring constant and x is the
displacement of the spring from its
unstrained length. The minus sign
indicates that the restoring force always
points in a direction opposite to the
displacement of the spring.

14. The SHM’s equations are as follows:
d2x It is the “equation of motion of the
∑ Fx = −kx ax =
dt 2 simple harmonic oscillator”. It is the
basis of many complex oscillator
d2x problems.
− kx = m
dt 2
d2x k
+ x =0
dt 2 m
The tentative solution of SHM’s equation is
x = x m cos(ωt + φ )
If it is differentiate twice with respect to the time
d2x
= −ω 2 x m cos(ωt + φ )
dt 2
Putting back into the original equation, obtained
k
− ω 2 x m cos(ωt + φ ) = − x m cos(ωt + φ )
m
Therefore, if we choose the constant ω such that
k
ω2 =
m
This is in fact a solution of the SHM’s equation.
The quantity ω is called the angular frequency, where
ω = 2π f
2π
and T =
ω
The quantity ωt + φ is called phase of the motion.
φ is called phase constant

15. xm, the maximum value of displacement, and φ are determined
by the initial position and velocity of the particles.
ω is determined by the system.
Oscillating Mass
Consider a mass m attached to
the end of a spring as shown. If
the mass is pulled down and
released, it will undergo simple
harmonic motion.
The period depends on the
spring constant, k and the mass
m, as given below,
m
T = 2π .
k
2
Therefore, m = T k
4π 2
Mass of an Astronaut
Astronauts who spend long periods of time in orbit periodically
measure their body masses as part of their health-maintenance
programs. On earth, it is simple to measure body mass, with a
scale.
However, this procedure does not work in orbit, because both the
scale and the astronaut are in free-fall and cannot press against
each other.

16. This device
consists of a
spring-mounted
chair in which
the astronaut
sits. The chair is
then started
oscillating in
simple harmonic
motion. The
period of the
motion is
measured
electronically
and is automatically converted into a value of the astronaut’s
mass, after the mass of the chair is taken into account.
How to understand φ ?
x = xm cos(ωt + ϕ )
x
x −t
xm ϕ =0
t π
o ϕ =
T 2
− xm ϕ =π

17. How to compare the phases of two SHOs with same ω?
x
Δϕ = 0
t
o
x
Δϕ = ± π
t
o
x
x1 = xm1 cos(ωt + ϕ1 )
x2 = x m2 cos(ωt + ϕ2 )
t
o
Δϕ = (ωt + ϕ2 ) − (ωt + ϕ1 )
Δϕ = ϕ2 − ϕ1

18. Displacement, Velocity, and Acceleration
Displacement, velocity and acceleration are the physical simple
harmonic motion descriptions similar to those in Kinematics. In
terms of mathematics, they are as follows:
Displacement x = x m cos(ωt + φ )
dx
Velocity v x = = −ωx m sin(ωt + φ )
dt
π
= ωx m cos(ωt + + φ)
2
d2x
Acceleration ax = = −ω 2 xm cos(ωt + φ )
dt 2
= ω 2 x m cos(ωt + π + φ )
When the displacement is a maximum in either direction, the
speed is zero, because the velocity must now change its
direction.

19. x
x − t graph
xm
x = x m cos(ωt + ϕ )
t
o
2π T
T = ϕ =0
ω
− xm
v
v−t graph
xmω
v = − x mω sin(ωt + ϕ )
T t
⎛ π⎞ o
= x mω cos⎜ ωt + ϕ + ⎟
⎝ 2⎠
− xmω
a a − t graph
xmω 2
a = − xmω 2 cos(ωt + ϕ ) t
o T
= x mω 2 cos(ωt + ϕ + π )
− xmω 2

20. SHM Parameters
A amplitude Maximum displacement right or left
T periode Time to do one oscillation, returning to start
f frequency Number of oscillations per second
1
T
PE potential Energy instantaneously stored in spring
energy
1
2 kx 2
KE kinetic Kinetic energy of motion of mass
energy
1
2 mv 2
E KE + PE Total mechanical energy in oscillations
vmax Maximum speed of the mass, as it passes x = 0.
ENERGY IN SIMPLE HARMONIC MOTION
The total energy can either be written using maximum x = A or
the maximum speed vmax:
2
E = 1
2 mv max = 1
2 kA2 = 1
2 mv 2 + 1
2 kx 2
when x = A,
v = 0, , all the
when v = vmax,
energy is PE
x = 0, all the
energy is KE
So we get a relation between vmax and A:
2
⎛ v max ⎞ k
⎜ ⎟ = a relation that shows how (k/m) influence
⎝ A ⎠ m
simple harmonic motion

21. Also get the velocity (or speed) as function of x:
⎛ A2 − x 2 ⎞ ⎛ x2 ⎞
v2 =
k
( )
A2 − x 2 = v max ⎜
2
⎜ A2 ⎟ = v max ⎜1 − 2 ⎟
⎟
2
⎜
m ⎝ ⎠ ⎝ A ⎟⎠
φ =0
1
0.8
0.6
0.4
0.2
T/2 T
1 2 3 4 5 6
1 1
The Potential Energy U =
2
kx 2 = kx m cos2 (ωt + φ )
2 2
The Kinetic Energy 1 1 2
K = mv 2 = mω 2 x m sin2 (ωt + φ )
2 2
1 2
= kx m sin2 (ωt + φ )
2
v = − xmω sin(ωt + ϕ )
Both Potential and Kinetic energies oscillate with time t and vary
between zero and maximum value of 12 k x m .2
Both Potential and Kinetic energies vary with twice the frequency
of the displacement and velocity

22. SHM Comparison to Circular Motion
v Look at x-component
y of velocity in circular
θ motion, radius A.
x
We can show that it
is analogous to the
A2 − x 2 velocity in SHM!
θ
x x vx = − vmax sin θ
= − vmax sin ωt
A2 − x 2 x2
But, sin θ = = 1− 2
A A
x2
Then, v x = − v max 1 −
A2
2π A
So, v max =
T
It shows that the projection of circular motion onto the x-axis is
the same as SHM along the x-axis.

23. SIMPLE PENDULUM
Figure at left shows a simple
pendulum of length L and particle
mass m
The restoring force is:
θ
T Fτ = −mg sin θ
L
If the θ is small, sin θ ≈ θ
m k x
θ
Fτ ≈ −mgθ = −mg = m&&
x
x L
mg
m m L
T = 2π = 2π = 2π
k mg / L g
Spring Configurations
In some circumstances, springs could be
configure as:
m
- single spring T = 2π
k
- series springs
1 1 1
= + ; T2 = 12 T1 (if k2 = 2k1 )
k ser k1 k2
- parallel springs
k par = k1 + k2 ; T2 = 1
2 T1 (if k2 = 2k1 )

24. Exercises
[1] A spring is hanging vertically, its initial length is 20 cm. It is
weighted by 100 grams mass so its length become 25 cm.
Calculate the potential energy of spring when it scretchs 10
cm long.
[2] An object is hanging onto vertical spring then being pulled
down 10 cm long and released. If the period is 0.2 s,
determine
(a) spring displacement after vibrates in 2 seconds
(b) the velocity of vibration after 2 s
(c) the acceleration of vibration after 2 s.
[3] In an experiment of simple pendulum, used a 1 m rope.
Time needed for 10 vibrations is 20 s. Determine the
gravitational acceleration of the experiment
[4] Base on vibration system on left picture, if k =
100 N and mass weight is 250 grams, determine
(a) period of vibration
(b) frequency of vibration
[5] A particle experiences a simple harmonic motion with
amplitude of 10 cm and periode of 4 s. After vibrates ¾T
determine
(a) displacement of vibration
(b) velocity of vibration
(c) acceleration of vibration