ENGGPHYS: FN-MODULE-1 1

ELASTICITY & OSCILLATIONS

ELASTICITY
Elasticity is the tendency of solid objects and materials to return to their original
shape after the external forces (load) causing a deformation are removed.
An elastic body is one that returns to its original form or shape after its deformation.
A rubber band going back to its original form after stretching, a soccer ball experiencing
deformation when it gets hit and soon return to its original shape after, and even a golf ball
are examples of an elastic body. An inelastic body meanwhile, is one that doesn’t return to
its original shape after experiencing deformation. Examples are dough and clay.

HOOKE’S LAW
Recall Hooke's law — first stated formally by Robert Hooke in The True Theory of
Elasticity or Springiness (1676)…
ut tension, sic vis
which can be translated literally to:
as extension, so force.
or can be translated formally to:
Extension is directly proportional to force.

Most likely we'd replace the word "extension" with the symbol (∆x), "force" with the
symbol (F), and "is directly proportional to" with an equals sign (=) and a constant of
proportionality (k), then, to show that the springy object was trying to return to its original
state, we'd add a negative sign (−). In other words, we'd write the equation…
𝑭 = −𝒌∆𝒙

A spring is an example of an elastic body that can be deformed

by stretching.
The force, which we can also call the restoring force, acts in the
direction opposite the displacement of the oscillating body. The spring
constant k is a measure of the elasticity of the spring:
𝑭
𝒌=
∆𝒙

STRESS & STRAIN

A change in shape due to the application of force is known as deformation. Even very
small forces are known to cause some deformation. Deformation is experienced by objects or
physical media under the action of external forces. In the language of physics, two terms
describe the forces on objects undergoing deformation: stress and strain.

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Stress is a quantity that describes the magnitude of forces that cause deformation. In
simplest statement, it refers to the cause of deformation. Stress is generally defined as force
per unit area.
𝑭𝒐𝒓𝒄𝒆, 𝑭
𝑺𝒕𝒓𝒆𝒔𝒔, 𝑺 =
𝑪𝒓𝒐𝒔𝒔 − 𝑺𝒆𝒄𝒕𝒊𝒐𝒏𝒂𝒍 𝑨𝒓𝒆𝒂, 𝑨
Types of Stress:
❖ Tensile Stress. A tensile stress occurs when equal and
opposite forces are directed away from each other. It is
when forces pull on an object and that force cause its
elongation.
❖ Compressive Stress. It occurs when equal and
opposite forces are directed toward each other.
❖ Bulk (Volume) Stress. It occurs when an object is
being squeezed from all sides.
❖ Shear Stress. In other situations, the acting forces
may be neither tensile nor compressive, and still
produce a noticeable deformation. In such a case,
when deforming forces act tangentially to the
object’s surface, we call them ‘shear’ forces and
the stress they cause is called shear stress.

The SI unit of stress is the pascal (Pa).

1 𝑁𝑒𝑤𝑡𝑜𝑛 (𝐹𝑜𝑟𝑐𝑒)
1 𝑃𝑎 (𝑠𝑡𝑟𝑒𝑠𝑠) =
1 𝑚2 (𝑎𝑟𝑒𝑎)
Other units:
1 𝑙𝑏
English: 1 𝑝𝑠𝑖 (𝑝𝑜𝑢𝑛𝑑 𝑝𝑒𝑟 𝑠𝑞𝑢𝑎𝑟𝑒 𝑖𝑛𝑐ℎ) = 1 𝑖𝑛2

❖ Elastic Limit
If the stress exceeds elastic
The elastic limit is the maximum stress a body
can experience without becoming permanently
limit, the final length will
deformed. be longer than the original
❖ Ultimate Strength
length (deformation)!
The ultimate strength is the greatest stress a
If the stress exceeds
body can experience without breaking or
rupturing.
ultimate strength, the
material breaks!

Strain is given as a fractional change in either length (under tensile stress) or volume
(under bulk stress) or geometry (under shear stress). In other words, strain refers to the effect
of the deformation.
𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒍𝒆𝒏𝒈𝒕𝒉 𝒐𝒓 𝒅𝒆𝒇𝒐𝒓𝒎𝒂𝒕𝒊𝒐𝒏 𝒍𝒆𝒏𝒈𝒕𝒉, ∆𝑳
𝑺𝒕𝒓𝒂𝒊𝒏, 𝜺 =
𝒊𝒏𝒊𝒕𝒊𝒂𝒍 𝒍𝒆𝒏𝒈𝒕𝒉, 𝑳

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Therefore, strain is a dimensionless number. Strain under a tensile stress is called

tensile strain, strain under bulk stress is called bulk strain (or volume strain), and that caused
by shear stress is called shear strain.

To summarize:
Stress = cause of deformation
Strain = effect of deformation

Sample Problems:
1. A steel wire 10 m long and 2 mm in diameter is attached to the ceiling. A 200-N weight
is attached to the end.
a. What is the applied stress?
b. If the 10 m steel wire stretches 3.08 mm due to the 200 N load. What is the
longitudinal strain?
c. The elastic limit for steel is 2.48 x 108 Pa. What is the maximum weight that can
be supported without permanently deforming the wire?
d. The ultimate strength for steel is for steel is 4.89 x 108 Pa. What is the maximum
weight supported without breaking the wire?

Given: L = 10 m
D = 2 mm
F = 200 N
Required: S = ?

𝐹𝑜𝑟𝑐𝑒
a) 𝑆= 𝐴𝑟𝑒𝑎
𝐴𝑟𝑒𝑎 = ?

𝜋𝐷 2
Solving for Area, we use: 𝐴 = = 𝜋𝑅 2
4
Note: formula is area of a circle. A circular section is shown when you cut a part of the steel wire
1𝑚
𝜋𝐷 2 𝜋[(2 𝑚𝑚)( )2
1000 𝑚𝑚
𝐴= 4
= 4
−𝟔
𝑨 = 𝟑. 𝟏𝟒𝟏𝟓𝟗 𝒙 𝟏𝟎 𝒎𝟐

𝐹
𝑆= 𝐴
200 𝑁
𝑆= 3.14159 𝑥 10−6 𝑚2
𝑺 = 𝟔. 𝟑𝟔𝟔 𝒙 𝟏𝟎𝟕 𝑷𝒂

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b)
Given: L = 10 m
△L = 3.08 mm
Req’d: strain

1𝑚
∆𝐿 3.08 𝑚𝑚 (1000𝑚𝑚)
𝜀 = =
𝐿 10 𝑚
−𝟒
𝜺 = 𝟑. 𝟎𝟖 𝒙 𝟏𝟎

c)
Given: s (elastic limit) = 2.48 x 108 Pa
Req’d: F = ?

𝐹
𝑠= 𝐴
𝐹
2.48 𝑥 108 𝑃𝑎 =
3.14159 𝑥 10−6 𝑚2
𝐹 = (2.48 𝑥 108 𝑃𝑎)(3.14159 𝑥 10−6 𝑚2 )
𝑭 = 𝟕𝟕𝟗. 𝟏𝟏𝟒 𝑵

d) Given: s (ultimate strength) = 4.89 x 108 Pa

Req’d: F = ?

𝐹
𝑠=
𝐴
𝐹
4.89 𝑥 108 𝑃𝑎 =
3.14159 𝑥 10−6 𝑚2
𝐹 = (4.89 𝑥 108 𝑃𝑎)(3.14159 𝑥 10−6 𝑚2 )
𝑭 = 𝟏𝟓𝟑𝟔. 𝟐𝟑𝟖 𝑵

2. A spring is stretched 0.115 m when a 2.0 N weight is hung from it. What is the spring
constant?

Given: x = 0.115 m
F = 2.0 N
Req’d: k = ?
𝐹
𝑘= 𝑥
2.0 𝑁
𝑘= 0.115 𝑚
𝒌 = 𝟏𝟕. 𝟒 𝑵/𝒎

3. A spring attached to a ceiling is hanging empty at 54.5 cm. When a 4 N object is

attached, it reaches to 83.4 cm. What is the spring constant?

Given: L = 54.5 cm
L after application of force = 83.4
F=4N
Req’d: k = ?

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𝐹
𝑘= ∆𝑥
4𝑁
𝑘= 1𝑚
(83.4 𝑐𝑚− 54.5 𝑐𝑚)( )
100 𝑐𝑚

𝒌 = 𝟏𝟑. 𝟖𝟒𝟏 𝑵/𝒎

WORKSHEET 1: STRESS & STRAIN

1. What is the force needed to stretch a spring, with a spring constant equal to 13 N/m, a
total of 0.16 m?
2. The elastic limit of steel is 5.0 Pa (N/m2), find the minimum diameter a steel wire can
to support a 700 N weight without exceeding its elastic limit.
3. Calculate the force a piano tuner applies to stretch a steel piano wire by 8.00 mm, if
the wire is originally 1.35 m long and its diameter is 0.850 mm.
4. A metal wire is 2.5 mm diameter and 2 m long. A force of 12 N is applied to it and it
stretches 0.3 mm. Assuming the metal is elastic, determine:
a. stress in the wire
b. strain in the wire
5. A 7 m long steel wire, 2.3 mm in diameter, is attached to a ceiling vertically and carries
a 50 kg object. Elastic limit = 3.6 x 108 Pa; Ultimate Strength = 7.2 x 108 Pa.
a. What is the applied stress?
b. What is the strain in the wire if it stretches 5mm due to the applied load?
c. What is the maximum load that can be carried by the wire without rupturing?
d. How great a load can be supported without permanently elongating the steel
wire?

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MODULUS OF ELASTICITY
The greater the stress, the greater the strain; however, the relation between strain and
stress does not need to be linear. Only when stress is sufficiently low is the deformation it
causes in direct proportion to the stress value. The proportionality constant in this relation is
called the elastic modulus or modulus of elasticity. In the linear limit of low stress values,
the general relation between stress and strain is
𝒔𝒕𝒓𝒆𝒔𝒔, 𝑺
𝒎𝒐𝒅𝒖𝒍𝒖𝒔 𝒐𝒇 𝒆𝒍𝒂𝒔𝒕𝒊𝒄𝒊𝒕𝒚 =
𝒔𝒕𝒓𝒂𝒊𝒏, 𝜺
❖ YOUNG’S MODULUS
For materials whose length is much greater than the width or thickness, we are
concerned with the longitudinal modulus of elasticity, or Young’s Modulus (Y).
Young’s modulus Y is the elastic modulus when deformation is caused by either
tensile or compressive stress.
𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠
𝑦𝑜𝑢𝑛𝑔′ 𝑠 𝑚𝑜𝑑𝑢𝑙𝑢𝑠, 𝑌 =
𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛
𝐹
𝑌= 𝐴
∆𝐿
𝐿
𝑭∙𝑳
𝒀=
𝑨 ∙ ∆𝑳

The SI unit for Young’s Modulus is Pa (pascal). English unit is lb/in2.

Sample Problems:
1. For safety in climbing, mountaineers use a nylon rope that is 50 m long and 0.01 m in
diameter. When supporting a 900 N climber, the rope stretches 1.6 m under tension.
Find the Young’s Modulus (Y) for the rope.

Given: L = 50 m
D = 0.01 m
F = 900 N
△L = 1.6 m
Req’d: Y = ?

𝐹𝐿 𝜋𝐷 2
𝑌= ; 𝐴=
𝐴∆𝐿 4
900 𝑁 (50 𝑚)
𝑌= 𝜋(0.01 𝑚)2
(1.6 𝑚)
4
𝟔
𝒀 = 𝟑𝟓𝟖. 𝟎𝟗𝟗 𝒙 𝟏𝟎 𝑷𝒂

2. A mass of 2.00 kg is supported by a copper wire of length 4.00 m and diameter 4.00
mm. Ycu = 1.6 x 106 Pa. Determine:

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a. the stress in the wire and

b. the elongation of the wire.

Given: m = 2.00 kg
L = 4.00 m
D = 4.00 mm

a) Req’d: S = ?

𝐹
𝑆= 𝐴
Note: convert mass into weight (force); using 9.81 m/s2 as the value of gravity
𝜋(𝐷)2
𝐹 = 𝑚𝑔; 𝐴 =
4
2.00 𝑘𝑔 (9.81 𝑚/𝑠2 )
𝑆= 1𝑚
𝜋(4.00 𝑚𝑚 𝑥 )2
1000 𝑚𝑚
4

𝑺 = 𝟏. 𝟓𝟔𝟏 𝒙 𝟏𝟎𝟔 𝑷𝒂

b) Req’d: △L = ?

𝐹𝐿
𝑌= 𝐴∆𝐿
𝑭𝑳
∆𝑳 = 𝑨𝒀
𝑚
2 𝑘𝑔 (9.81 2 )(4.00 𝑚)
𝑠
∆𝐿 = 𝜋(0.004 𝑚)2
(1.6 𝑥 106 𝑃𝑎)
4
∆𝑳 = 𝟑. 𝟗𝟎𝟑 𝒎

3. A 5 kg mass hangs on a vertical steel wire 0.5 m long and 0.004 cm2 in cross-section.
Hanging from the bottom of this mass is a copper wire of the same length and 0.003
cm2 that supports a 10 kg mass. For each wire, compute:
a. the elongation, and
b. the longitudinal strain.

Given:
STEEL
L = 0.5 m
A = 0.004 cm2
Ys = 2 x 1011 Pa
5 KG

COPPER
L = 0.5 m
A = 0.003 cm2
Ycu = 1.1 x 1011 Pa
10 KG

a) Req’d: △LS (elongation of steel)

△Lcu (elongation of copper)

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STEEL:
𝐹𝑠 = (5 𝑘𝑔 + 10 𝑘𝑔) (9.81 𝑚/𝑠 2 )
*Note: The steel wire carries the two mass, therefore we consider them both as the applied force at steel wire
𝑭𝒔 = 𝟏𝟒𝟕. 𝟏𝟓 𝑵

1𝑚
𝐴𝑠 = 0.004 𝑐𝑚2 ( )2
100 𝑐𝑚
𝑨𝒔 = 𝟒 𝒙 𝟏𝟎−𝟕 𝒎 𝟐

Solving for the elongation:

147.15 𝑁 (0.5 𝑚)
∆𝐿𝑠 =
4 𝑥 10−7 𝑚2 (2 𝑥 1011 𝑃𝑎)

∆𝑳𝒔 = 𝟗. 𝟏𝟗𝟕 𝒙 𝟏𝟎−𝟒 𝒎

COPPER:
𝐹𝑐𝑢 = 10 𝑘𝑔 (9.81 𝑚/𝑠 2 )
𝑭𝒄𝒖 = 𝟗𝟖. 𝟏 𝑵

1𝑚
𝐴𝑐𝑢 = 0.003 𝑐𝑚2 ( )2
100 𝑐𝑚
𝑨𝒄𝒖 = 𝟑 𝒙 𝟏𝟎−𝟕 𝒎 𝟐

Solving for the elongation:

98.1 𝑁 (0.5 𝑚)
∆𝐿𝑠 =
3 𝑥 10−7 𝑚2 (1.1 𝑥 1011 𝑃𝑎)

∆𝑳𝒔 = 𝟗. 𝟏𝟗𝟕 𝒙 𝟏𝟎−𝟒 𝒎

b) Req’d: ɛs (strain @ steel wire)

ɛcu (strain @ copper wire)

STEEL:
𝑆
𝑌𝑠 = 𝜀
𝐹
𝑆 𝐴
𝜀𝑠 = =
𝑌 𝑌
𝐹
𝜀𝑠 = 𝐴𝑌
147.15 𝑁
𝜀𝑠 = 4 𝑥 10−7 𝑚2 (2 𝑥 1011 𝑃𝑎)
𝜺𝒔 = 𝟏. 𝟖𝟑𝟗 𝒙 𝟏𝟎−𝟑

COPPER:
𝐹
𝜀𝑠 = 𝐴𝑌

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98.1 𝑁
𝜀𝑠 = 3 𝑥 10−7 𝑚2 (1.1 𝑥 1011 𝑃𝑎)
𝜺𝒔 = 𝟐. 𝟗𝟕𝟑 𝒙 𝟏𝟎−𝟑

4. A copper rod of length 2 m and cross sectional area 2.0 cm2 is fastened end-to-end to
a steel rod of length L and cross-sectional area 1.0 cm2. The compund rod is subjected
to equal and opposite pulls of magnitude 3 x 104 N at its ends. Ycu = 1.1 x 1011 Pa; Ys =
2 x 1011 Pa.
a. Find the length L of the steel rod if the elongations of the two rods are equal.
b. What is the stress in each rod?
c. What is the strain in each rod?

Given:
Acu = 2.0 cm2
As = 1.0 cm2

F = 3 x 104 N COPPER STEEL F = 3 x 104 N

2m L

a) Req’d: L of steel rod

Condition:
∆𝐿𝑐𝑢 = ∆𝐿𝑠

𝐹𝐿 𝐹𝐿
(𝐴𝑌)𝑐𝑢 = (𝐴𝑌)𝑠
3 𝑥 104 𝑁(2 𝑚) 3 𝑥 104 𝑁(𝐿)
1𝑚 2 = 1𝑚 2
2.0 𝑐𝑚2 ( ) (1.1 𝑥 1011 𝑃𝑎) 1.0 𝑐𝑚2 ( ) (2 𝑥 1011 𝑃𝑎)
100 𝑐𝑚 100 𝑐𝑚

Isolating variable L on one side:

1𝑚 2
3 𝑥 104 𝑁(2 𝑚)(1.0 𝑐𝑚2 ( ) (2 𝑥 1011 𝑃𝑎)
100 𝑐𝑚
1 𝑚 =𝐿
2.0 𝑐𝑚2 ( )2 (1.1 𝑥 1011 𝑃𝑎)(3 𝑥 104 𝑁)
100 𝑐𝑚

𝑳 = 𝟏. 𝟖𝟏𝟖 𝒎

b) Req’d: Scu
Ss

COPPER ROD:
𝐹 3 𝑥 104 𝑁
𝑆𝑐𝑢 = = 1𝑚 2
𝐴 2.0 𝑐𝑚2 (100 𝑐𝑚)

𝑺𝒄𝒖 = 𝟏. 𝟓 𝒙 𝟏𝟎𝟖 𝑷𝒂

STEEL ROD:

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𝐹 3 𝑥 104 𝑁
𝑆𝑠 = = 1𝑚 2
𝐴 1.0 𝑐𝑚2 (100 𝑐𝑚)

𝑺𝒄𝒖 = 𝟑 𝒙 𝟏𝟎𝟖 𝑷𝒂

c) Req’d: ɛcu
ɛs

COPPER ROD:
∆𝐿
𝜀𝑐𝑢 = 𝐿
, ∆𝐿 = ?

*Solving for △L:

𝐹𝐿
∆𝐿𝑐𝑢 = ( )𝑐𝑢
𝐴𝑌
3 𝑥 104 𝑁(2 𝑚)
∆𝐿𝑐𝑢 = 1𝑚 2
2.0 𝑐𝑚2 (100 𝑐𝑚) (1.1 𝑥 1011 𝑃𝑎)
∆𝑳𝒄𝒖 = 𝟐. 𝟕𝟐𝟕 𝒙 𝟏𝟎−𝟑 𝒎

∆𝐿 2.727 𝑥 10−3 𝑚
𝜀𝑐𝑢 = =
𝐿 2𝑚
𝜺𝒄𝒖 = 𝟏. 𝟑𝟔𝟒 𝒙 𝟏𝟎−𝟑

STEEL ROD:
∆𝐿𝑠 = ∆𝐿𝑐𝑢 = 2.727 𝑥 10−3 𝑚

∆𝐿 2.727 𝑥 10−3 𝑚
𝜀𝑆 = =
𝐿 1.818 𝑚

𝜺𝑺 = 𝟏. 𝟓 𝒙 𝟏𝟎−𝟑

WORKSHEET 2: MODULUS OF ELASTICITY

1. A vertical beam in a building supports a load of 6.0 x 104 N. If the length of the beam is
4.0 m and its cross-sectional area is a circle with a diameter of 0.4 m, find the distance
the beam is compressed along its length. (Y = 2.7 x 108 Pa)
2. A 90-kg mountain climber hangs from a nylon rope and stretches it by 25.0 cm. If the
rope was originally 30.0 m long and its diameter is 1.0cm, what is Young’s modulus for
the nylon?
3. A cylindrical aluminum pillar 7.00 high has a radius of 30.0 cm. If a 2000-kg sculpture
is placed on top of the pillar, by how much is the pillar compressed? (Y = 7 x 1010 Pa)
4. A 7.5 kg mass hangs on a vertical steel wire (Y = 2 x 1011 Pa) 1m long and 0.005 cm2 in
cross-section. Hanging from the bottom of this mass is a similar steel wire, but 1.5 m in
length, that supports a 13 kg mass. For each wire,
a. compute the longitudinal strain

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b. and the elongation.

5. A copper wire 8 m long and a steel wire 4 m long, each of cross-section 0.5 cm2 and
0.3 cm2 respectively, are fastened end-to-end and stretched with a tension of 500 N.
Young’s modulus for copper is 1.1 x 1011 Pa; Young’s modulus for steel is 2 x 1011 Pa.
a. What is the stress in each wire?
b. What is the change in length of each wire?
c. What is the strain in each wire?

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