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Elasticity

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  • 1. ELASTICITY ANDHARMONIC MOTIONStandard CompetencyAnalyzes the nature phenomenon and its regularity within thescope of particle’s MechanicsBase CompetencyAnalyzes the effect of force on the elasticity properties of amaterialLearning Objectives1 Describes the characteristic of force on elastic material base on experiment performed2 Identifies the elastic moduli and spring force constant3 Compares the force constant base on observatiob’s data4 Analyses the series and parallel spring configuration5 Calculates spring’s elongation6 Determines the value of spring’s force constantReferences[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. ELASTICITYElasticity is the property or ability of an object or material torestored to its original shape after apllied distortion vanished.Elasticity ≠ PlasticsA spring is an example of an elastic object - when stretched, itexerts a restoring force which tends to bring it back to its originallength. This restoring force is generally proportional to theamount of stretch, as described by Hookes Law. For wires orcolumns, the elasticity is generally described in terms of theamount of deformation (strain) resulting from a given stress(Youngs modulus). Bulk elastic properties of materials describethe 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 isproportional to the object elongation ΔL ” F = −k ΔL(−) minus sign shows that the restoring force F is oppose toobject’s directionSTRESS (TENSION)Stress or tension is defined as force per unit cross section area.It has unit (SI) N/m2. F σ = ASTRAIN (SCRETCH)Stress or tension is defined as ratio between elongation andinitial length. It is unitless. Δl e= lo
  • 4. YOUNG’s MODULUSYoung’s modulus or elastic modulus is ration between stress andstrain. It has unit (SI) N/m2. σ 1 F E = → ΔL = L e E AE is the elastic modulus or Youngs modulus and is onlydependent 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 9Liquids Water 2.0 x 10 9 Alcohol 1.0 x 10 9(ethyl) Mercury 2.5 x 10 9Gases Air, He, H2, 1.01 x 10 5CO2
  • 5. How force is affecting a material in term of its elongationdescribed 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. ExampleA metal (steel) rod whose has cross section area of 4 mm2 andlength of 40 cm is hang and pulled down by force of 100 N. Ifthe elastic modulus of metal is 2 x 1011 N/m2, calculate(a) stress(b) strain(c) elongation lengthKnown A = 4 mm2 = 4 x 106 m2 lo = 40 cm = 0.4 m F = 100 N E = 2 x 104 N/m2Asked (a) stress, σ (b) strain, e (c) elongation length, ΔlAnswer 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 bathyspheres 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 ANDHARMONIC MOTIONStandard CompetencyAnalyzes the nature phenomenon and its regularity within thescope of particle’s MechanicsBase CompetencyAnalyzes the relation between force and harmonic motionLearning Objectives1 Describes the characteristic of motion on vibrate spring2 Explains the relation between the period of harmonic motion and mass weighted base on observation’s data3 Analyzes the displacement, velocity and acceleration planetary motion within a universe base on Keppler’s Law4 Analyses the potentian and mximum kinetic energy on harmonic motionReferences[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 asswinging pendulum of a clock, a person bouncing on atrampoline, 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 theequilibrium position, proportional to the displacement.Look at a mass on a spring as example x = displacement F = force due to spring F = −k xThe restoring force is opposite to the displacement.
  • 12. Give m a positive displacement where x = A. then release it. Fwill 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 sineand cosine functions is called a “simple harmonic oscillator” andits motion is called “simple harmonic motion”.Equation of motion of the simple harmonic oscillatorFigure in the right showsa simple harmonicoscillator, consisting of aspring of force constant Kacting on a body of massm that slides on africtionless horizontalsurface. The body movesin 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 mThe 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 2Putting back into the original equation, obtained k − ω 2 x m cos(ωt + φ ) = − x m cos(ωt + φ ) mTherefore, if we choose the constant ω such that k ω2 = mThis is in fact a solution of the SHM’s equation.The quantity ω is called the angular frequency, where ω = 2π f 2πand T = ωThe quantity ωt + φ is called phase of the motion.φ is called phase constant
  • 15. xm, the maximum value of displacement, and φ are determinedby the initial position and velocity of the particles.ω is determined by the system.Oscillating MassConsider a mass m attached tothe end of a spring as shown. Ifthe mass is pulled down andreleased, it will undergo simpleharmonic motion.The period depends on thespring constant, k and the massm, as given below, m T = 2π . k 2Therefore, m = T k 4π 2Mass of an AstronautAstronauts who spend long periods of time in orbit periodicallymeasure their body masses as part of their health-maintenanceprograms. On earth, it is simple to measure body mass, with ascale.However, this procedure does not work in orbit, because both thescale and the astronaut are in free-fall and cannot press againsteach 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 electronicallyand is automatically converted into a value of the astronaut’smass, 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 ) to Δϕ = (ωt + ϕ2 ) − (ωt + ϕ1 ) Δϕ = ϕ2 − ϕ1
  • 18. Displacement, Velocity, and AccelerationDisplacement, velocity and acceleration are the physical simpleharmonic motion descriptions similar to those in Kinematics. Interms of mathematics, they are as follows:Displacement x = x m cos(ωt + φ ) dxVelocity v x = = −ωx m sin(ωt + φ ) dt π = ωx m cos(ωt + + φ) 2 d2xAcceleration 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, thespeed is zero, because the velocity must now change itsdirection.
  • 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 xmωv = − 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 MOTIONThe total energy can either be written using maximum x = A orthe 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 KESo 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 1The Potential Energy U = 2 kx 2 = kx m cos2 (ωt + φ ) 2 2The 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 varybetween zero and maximum value of 12 k x m .2Both Potential and Kinetic energies vary with twice the frequencyof 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 x2But, sin θ = = 1− 2 A A x2Then, v x = − v max 1 − A2 2π ASo, v max = TIt shows that the projection of circular motion onto the x-axis isthe 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 gSpring 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