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- ...
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 2 3 4 5 6
Describes the characteristic of force on elastic material base on experiment performed Identifies the elastic moduli and spring force constant Compares the force constant base on observatiob’s data Analyses the series and parallel spring configuration Calculates spring’s elongation 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
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.
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.
σ =
F A
STRAIN (SCRETCH) Stress or tension is defined as ratio between elongation and initial length. It is unitless.
e=
Δl lo
YOUNG’s MODULUS Young’s modulus or elastic modulus is ration between stress and strain. It has unit (SI) N/m2.
E =
σ e
→
ΔL =
1 F L 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 Solids Steel Brass Aluminum Concrete Brick Bone (limb) Liquids Water Alcohol (ethyl) Mercury Gases Air, He, H2, CO2
Elastic Modulus E ( N/m2 ) 200 x 10 9 100 x 10 9 70 x 10 9 20 x 10 9 14 x 10 9 15 x 10 9
Shear Modulus G ( N/m2 ) 80 x 10 9 35 x 10 9 25 x 10 9
Bulk Modulus B ( N/m2 ) 140 x 10 9 80 x 10 9 70 x 10 9
80 x 10 9 2.0 x 10 9 1.0 x 10 9 2.5 x 10 9 1.01 x 10 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
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
(a) σ =
F 100 = = 2.5 x 10− 6 N/m2 −6 A 4 x 10
σ
2.5 x 107 = = 1.25 x 10− 4 (b) E = 11 e 2 x 10 (c) Δl = e x lo = (1.25 x 10−4) (4 x 104) = 5 x 105 m
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?
Answers [1]
ΔL =
1 F 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. ΔL = 0.37 cm r = 0.00425 m Lo = 15 cm 2 2 A = r = ( 0.00425 m ) = 5.7 x 10 -5 m2 ΔL =
[2]
1 F 1 13.4 N Lo = 15 cm = 9.5 . 106 N/m2 −5 2 E A 0.37 cm 5.7 . 10 m
ΔV 1 = − ΔP Vo B ΔP = −B
solve for ΔP
ΔV − 0.10 = − 90 . 109 N/m2 = 9.0 . 107 N/m2 100 Vo
9.0 . 107 N/m2 = 900 1.0 . 105 N/m2 It is 900 times greater than atmospheric pressure
[3]
ΔV 1 1 = − ΔP = − 2.0 . 107 N/m2 = − 2.2 . 10 − 4 9 2 Vo B 90 . 10 N/m Percentage = - 2.2 x 10 -4 x 100 = - 2.2 x 10 -2
[4]
A = πr2 = 3.14 (0.0005 m)2 = 7.9 x 10 -7 m2 1 F 1 250 N Lo = 30.0 cm = 1.9 cm 9 -2 E A 5 . 10 Nm 7.9 . 10- 7 m2
ΔL =
[5] F = mg = 1550 kg x 9.80 m s -2 = 15200 N
stress =
F 15200 = = 1.0 . 105 N/m2 A 0.15
ΔL 1 1 = stress = 1.0 . 105 N/m2 = 5.0 . 10 − 7 9 2 Lo E 200 . 10 N/m
strain =
ΔL = Strain x Lo = 5.0 x 10 -7 x 9.50 m = 4.8 x 10 -6 m
[6]
ΔL =
1 F Lo E A
PE =
1
2
k ΔL2
1 F F Lo = E A k k =
and
F = kΔL ⇒
ΔL =
F k
set ΔL = ΔL solved for k ; 1 m = 100 cm, 1 m2 = 1000 cm2
E A 2.0 . 10 6 N/m2 x 5.0 . 10-5 m2 = = 3.3 . 10 4 N/m -3 Lo 3.0 . 10 m
PE = ½ k ΔL2 = ½ (3.3 . 104 N/m)(1.0 . 10−3 m)2 = 0.017 J
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 2 3 4
Describes the characteristic of motion on vibrate spring Explains the relation between the period of harmonic motion and mass weighted base on observation’s data Analyzes the displacement, velocity and acceleration planetary motion within a universe base on Keppler’s Law 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
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.
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.
f =
1 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.
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.
The SHM’s equations are as follows:
∑ Fx = −kx − kx = m
ax =
d2x dt 2
d2x dt 2
It is the “equation of motion of the simple harmonic oscillator”. It is the basis of many complex oscillator problems.
d2x k + x =0 m dt 2
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 − ω 2 x m cos(ωt + φ ) = −
k x m cos(ωt + φ ) m
Therefore, if we choose the constant ω2 =
ω such that
k m
This is in fact a solution of the SHM’s equation. The quantity ω is called the angular frequency, where ω = 2π f
and T =
2π
ω
The quantity ωt + φ is called phase of the motion. φ is called phase constant
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, T = 2π
m . k
2 Therefore, m = T k2 4π
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.
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
o
− xm
ϕ =0
t
T
ϕ =
π 2
ϕ =π
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 )
o
t
Δϕ = (ωt + ϕ2 ) − (ωt + ϕ1 ) Δϕ = ϕ2 − ϕ1
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 )
Velocity v x =
dx = −ωx m sin(ωt + φ ) dt = ωx m cos(ωt +
Acceleration ax =
π 2
+ φ)
d2x = −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.
x xm
x − t graph
x = x m cos(ωt + ϕ )
T =
2π
ω
t
o
T
ϕ =0 − xm v
xmω
v−t
graph
v = − x mω sin(ωt + ϕ )
π⎞ ⎛ = x mω cos⎜ ωt + ϕ + ⎟ 2⎠ ⎝
o
T
t
T
t
− xmω a
a − t graph
xmω 2
a = − xmω 2 cos(ωt + ϕ ) = x mω 2 cos(ωt + ϕ + π )
o
− xmω 2
SHM Parameters A T f
amplitude periode frequency
Maximum displacement right or left Time to do one oscillation, returning to start Number of oscillations per second
1 T
PE
potential energy 1
KE
vmax
kx 2
kinetic energy 1
E
2
Energy instantaneously stored in spring
2
Kinetic energy of motion of mass
mv 2
KE + PE Total mechanical energy in oscillations 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: E =
1
2
2 mv max =
1
2
kA2 =
1
2
mv 2 +
when v = vmax, x = 0, all the energy is KE
1
2
kx 2
when x = A, v = 0, , all the energy is PE
So we get a relation between vmax and A: 2
k ⎛ v max ⎞ ⎜ ⎟ = m ⎝ A ⎠
Æ a relation that shows how (k/m) influence simple harmonic motion
Also get the velocity (or speed) as function of x: v2 =
⎛ A2 − x 2 ⎞ ⎛ k x2 ⎞ 2 2 ⎜⎜ ⎟ ⎜ ⎟ = − A2 − x 2 = v max v 1 max ⎜ 2 ⎟ m A2 ⎟⎠ ⎝ A ⎠ ⎝
(
)
φ =0 1 0.8 0.6 0.4 0.2
T/2 1
The Potential Energy
The Kinetic Energy
2
3
U =
T 4
5
6
1 1 2 kx 2 = kx m cos2 (ωt + φ ) 2 2
1 1 2 mv 2 = mω 2 x m sin2 (ωt + φ ) 2 2 1 2 = kx m sin2 (ωt + φ ) 2
K =
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 m2 . Both Potential and Kinetic energies vary with twice the frequency of the displacement and velocity
SHM Comparison to Circular Motion v
y
x
θ
A2 − x 2
θ
x
x
Look at x-component of velocity in circular motion, radius A. We can show that it is analogous to the velocity in SHM! vx = − vmax sin θ = − vmax sin ωt
A2 − x 2 x2 = 1− 2 A A
But,
sin θ =
Then,
v x = − v max 1 −
So,
v max =
x2 A2
2π A T
It shows that the projection of circular motion onto the x-axis is the same as SHM along the x-axis.
SIMPLE PENDULUM Figure at left shows a simple pendulum of length L and particle mass m The restoring force is:
θ
Fτ = −mg sin θ
T
L
If the θ is small, sin θ ≈ θ m
x
θ
k
Fτ ≈ −mgθ = −mg
mg
T = 2π
x = m&x& L
m m L = 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 )
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
