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Chapter 10 Elasticity & Oscillations Elasticity and Oscillations • Elastic Deformations • Hooke’s Law • Stress and Strain • Shear Deformations • Volume Deformations • Simple Harmonic Motion • The Pendulum • Damped Oscillations, Forced Oscillations, and Resonance MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 2 Elastic Deformation of Solids A deformation is the change in size or shape of an object. An elastic object is one that returns to its original size and shape after contact forces have been removed. If the forces acting on the object are too large, the object can be permanently distorted. MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 3 Hooke’s Law F F Apply a force to both ends of a long wire. These forces will stretch the wire from length L to L+L. MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 4 Stress and Strain Define: L strain L The fractional change in length F stress A Force per unit crosssectional area Stretching ==> Tensile Stress Squeezing ==> Compressive Stress MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 5 Hooke’s Law Hooke’s Law (Fx) can be written in terms of stress and strain (stress strain). F L Y A L The spring constant k is now YA k L Y is called Young’s modulus and is a measure of an object’s stiffness. Hooke’s Law holds for an object to a point called the proportional limit. MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 6 Compressive Stress A steel beam is placed vertically in the basement of a building to keep the floor above from sagging. The load on the beam is 5.8104 N and the length of the beam is 2.5 m, and the cross-sectional area of the beam is 7.5103 m2. Find the vertical compression of the beam. Force of ceiling on beam Force of floor on beam F L Y A L F L L A Y For steel Y = 200109 Pa. 4 2.5 m F L 5.8 10 N 4 L 1 . 0 10 m 3 2 9 2 A Y 7.5 10 m 200 10 N/m MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 7 Tensile Stress Example (text problem 10.7): A 0.50 m long guitar string, of cross-sectional area 1.0106 m2, has a Young’s modulus of 2.0109 Pa. By how much must you stretch a guitar string to obtain a tension of 20 N? F L Y A L 0.5 m F L 20.0 N L 6 2 9 2 A Y 1.0 10 m 2.0 10 N/m 5.0 10 3 m 5.0 mm MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 8 Beyond Hooke’s Law Elastic Limit If the stress on an object exceeds the elastic limit, then the object will not return to its original length. Breaking Point An object will fracture if the stress exceeds the breaking point. The ratio of maximum load to the original cross-sectional area is called tensile strength. MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 9 MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 10 Ultimate Strength The ultimate strength of a material is the maximum stress that it can withstand before breaking. Materials support compressive stress better than tensile stress MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 11 An acrobat of mass 55 kg is going to hang by her teeth from a steel wire and she does not want the wire to stretch beyond its elastic limit. The elastic limit for the wire is 2.5108 Pa. What is the minimum diameter the wire should have to support her? F stress elastic limit A Want F mg A elastic limit elastic limit 2 mg D elastic limit 2 D MFMcGraw-PHY1401 4mg 1.7 10 3 m 1.7 mm elastic limit Chap 10d - Elas & Vibrations - Revised 7-12-10 12 Different Representations F ΔL =Y A L0 ΔL 1 F = L0 Y A Original equation Fractional change 1 F ΔL = L0 Y A 1 F L - L0 = L0 Y A 1 F L = L0 + L0 Y A 1 F L = L0 1+ Y A MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 Change in length New length 13 Shear Deformations A shear deformation occurs when two forces are applied on opposite surfaces of an object. MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 14 Stress and Strain Define: Shear Force F Shear Stress Surface Area A displaceme nt of surfaces x Shear Strain separation of surfaces L Hooke’s law (stressstrain) for shear deformations is F x S A L MFMcGraw-PHY1401 where S is the shear modulus Chap 10d - Elas & Vibrations - Revised 7-12-10 15 Example (text problem 10.25): The upper surface of a cube of gelatin, 5.0 cm on a side, is displaced by 0.64 cm by a tangential force. The shear modulus of the gelatin is 940 Pa. What is the magnitude of the tangential force? F F x S A L F From Hooke’s Law: x F SA L 0.64 cm 940 N/m 0.0025 m 0.30 N 5.0 cm MFMcGraw-PHY1401 2 2 Chap 10d - Elas & Vibrations - Revised 7-12-10 16 Volume Deformations An object completely submerged in a fluid will be squeezed on all sides. F volume stress pressure A V The result is a volume strain; volume strain V MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 17 Volume Deformations For a volume deformation, Hooke’s Law is (stressstrain): V P B V where B is called the bulk modulus. The bulk modulus is a measure of how easy a material is to compress. MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 18 An anchor, made of cast iron of bulk modulus 60.0109 Pa and a volume of 0.230 m3, is lowered over the side of a ship to the bottom of the harbor where the pressure is greater than sea level pressure by 1.75106 Pa. Find the change in the volume of the anchor. V P B V VP 0.230 m 3 1.75 106 Pa V B 60.0 109 Pa 6.7110 6 m 3 MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 19 Examples • I-beam • Arch - Keystone • Flying Buttress MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 20 Deformations Summary Table Tensile or compressive Shear Volume Stress Force per unit cross-sectional area Shear force divided by the area of the surface on which it acts Pressure Strain Fractional change in length Ratio of the relative displacement to the separation of the two parallel surfaces Fractional change in volume Constant of Young’s modulus Shear modulus (S) proportionality (Y) MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 Bulk Modulus (B) 21 Simple Harmonic Motion Simple harmonic motion (SHM) occurs when the restoring force (the force directed toward a stable equilibrium point) is proportional to the displacement from equilibrium. MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 22 Characteristics of SHM • Repetitive motion through a central equilibrium point. • Symmetry of maximum displacement. • Period of each cycle is constant. • Force causing the motion is directed toward the equilibrium point (minus sign). • F directly proportional to the displacement from equilibrium. Acceleration = - ω2 x Displacement MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 23 The Spring The motion of a mass on a spring is an example of SHM. Equilibrium position y x x The restoring force is F = kx. MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 24 Equation of Motion & Energy Assuming the table is frictionless: F Classic form for SHM Also, MFMcGraw-PHY1401 x = - kx = ma x k ax t = - x t = - ω2 x t m 1 2 1 2 E t K t U t mv t kx t 2 2 Chap 10d - Elas & Vibrations - Revised 7-12-10 25 Spring Potential Energy MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 26 Simple Harmonic Motion At the equilibrium point x = 0 so a = 0 too. When the stretch is a maximum, a will be a maximum too. The velocity at the end points will be zero, and it is a maximum at the equilibrium point. MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 27 MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 28 What About Gravity? When a mass-spring system is oriented vertically, it will exhibit SHM with the same period and frequency as a horizontally placed system. The effect of gravity is cancelled out. MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 29 Spring Compensates for Gravity MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 30 Representing Simple Harmonic Motion MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 31 A simple harmonic oscillator can be described mathematically by: xt A cos t x vt A sin t t v at A 2 cos t t Or by: xt A sin t x vt A cos t t v at A 2 sin t t MFMcGraw-PHY1401 where A is the amplitude of the motion, the maximum displacement from equilibrium, A = vmax, and A2 = amax. Chap 10d - Elas & Vibrations - Revised 7-12-10 32 Linear Motion - Circular Functions MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 33 Projection of Circular Motion MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 34 The Period and the Angular Frequency The period of oscillation is where is the angular frequency of the oscillations, k is the spring constant and m is the mass of the block. MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 T 2 . k m 35 The period of oscillation of an object in an ideal mass-spring system is 0.50 sec and the amplitude is 5.0 cm. What is the speed at the equilibrium point? At equilibrium x = 0: 1 2 1 2 1 2 E K U mv kx mv 2 2 2 Since E = constant, at equilibrium (x = 0) the KE must be a maximum. Here v = vmax = A. MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 36 Example continued: The amplitude A is given, but is not. 2 2 12.6 rads/sec T 0.50 s and v Aω 5.0 cm 12.6 rads/sec 62.8 cm/sec MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 37 The diaphragm of a speaker has a mass of 50.0 g and responds to a signal of 2.0 kHz by moving back and forth with an amplitude of 1.8104 m at that frequency. (a) What is the maximum force acting on the diaphragm? 2 2 F F ma m A mA 2 f 4 mAf max max 2 2 The value is Fmax=1400 N. MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 38 Example continued: (b) What is the mechanical energy of the diaphragm? Since mechanical energy is conserved, E = Kmax = Umax. U max K max 1 2 kA 2 1 2 mvmax 2 The value of k is unknown so use Kmax. K max 1 2 1 1 2 2 mvmax m A mA2 2f 2 2 2 The value is Kmax= 0.13 J. MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 39 Example (text problem 10.47): The displacement of an object in SHM is given by: yt 8.00 cmsin 1.57 rads/sec t What is the frequency of the oscillations? Comparing to y(t) = A sint gives A = 8.00 cm and = 1.57 rads/sec. The frequency is: 1.57 rads/sec f 0.250 Hz 2 2 MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 40 Example continued: Other quantities can also be determined: The period of the motion is 2 2 T 4.00 sec 1.57 rads/sec xmax A 8.00 cm vmax A 8.00 cm 1.57 rads/sec 12.6 cm/sec amax A 2 8.00 cm 1.57 rads/sec 19.7 cm/sec 2 2 MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 41 The Pendulum A simple pendulum is constructed by attaching a mass to a thin rod or a light string. We will also assume that the amplitude of the oscillations is small. MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 42 The pendulum is best described using polar coordinates. The origin is at the pivot point. The coordinates are (r, φ). The r-coordinate points from the origin along the rod. The φcoordinate is perpendicualr to the rod and is positive in the counterclock wise direction. MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 43 2nd Apply Newton’s Law to the pendulum bob. F mg sin ma 2 v Fr T mg cos m r 0 If we assume that φ <<1 rad, then sin φ φ and cos φ 1, the angular frequency of oscillations is then: F mg sin ma mL mg sin mL ( g / L)sin ( g / L) The period of oscillations is MFMcGraw-PHY1401 T Chap 10d - Elas & Vibrations - Revised 7-12-10 g L 2 2 L g 44 Example (text problem 10.60): A clock has a pendulum that performs one full swing every 1.0 sec. The object at the end of the string weighs 10.0 N. What is the length of the pendulum? L T 2 g Solving for L: MFMcGraw-PHY1401 gT 2 9.8 m/s 2 1.0 s L 0.25 m 2 2 4 4 Chap 10d - Elas & Vibrations - Revised 7-12-10 2 45 The gravitational potential energy of a pendulum is U = mgy. Taking y = 0 at the lowest point of the swing, show that y = L(1-cos). Lcos L L y L(1 cos ) y=0 MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 46 The Physical Pendulum A physical pendulum is any rigid object that is free to oscillate about some fixed axis. The period of oscillation of a physical pendulum is not necessarily the same as that of a simple pendulum. MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 47 The Physical Pendulum http://hyperphysics.phy-astr.gsu.edu/HBASE/pendp.html MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 48 Damped Oscillations When dissipative forces such as friction are not negligible, the amplitude of oscillations will decrease with time. The oscillations are damped. MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 49 Graphical representations of damped oscillations: MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 50 Damped Oscillations Overdamped: The system returns to equilibrium without oscillating. Larger values of the damping the return to equilibrium slower. Critically damped : The system returns to equilibrium as quickly as possible without oscillating. This is often desired for the damping of systems such as doors. Underdamped : The system oscillates (with a slightly different frequency than the undamped case) with the amplitude gradually decreasing to zero. Source: Damping @ Wikipedia MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 51 Damped Oscillations The larger the damping the more difficult it is to assign a frequency to the oscillation. MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 52 Forced Oscillations and Resonance A force can be applied periodically to a damped oscillator (a forced oscillation). When the force is applied at the natural frequency of the system, the amplitude of the oscillations will be a maximum. This condition is called resonance. MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 53 Tacoma Narrows Bridge Nov. 7, 1940 MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 54 Tacoma Narrows Bridge Nov. 7, 1940 MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 55 Tacoma Narrows Bridge The first Tacoma Narrows Bridge opened to traffic on July 1, 1940. It collapsed four months later on November 7, 1940, at 11:00 AM (Pacific time) due to a physical phenomenon known as aeroelastic flutter caused by a 67 kilometres per hour (42 mph) wind. The bridge collapse had lasting effects on science and engineering. In many undergraduate physics texts the event is presented as an example of elementary forced resonance with the wind providing an external periodic frequency that matched the natural structural frequency (even though the real cause of the bridge's failure was aeroelastic flutter[1]). Its failure also boosted research in the field of bridge aerodynamics/ aeroelastics which have themselves influenced the designs of all the world's great long-span bridges built since 1940. - Wikipedia http://www.youtube.com/watch?v=3mclp9QmCGs MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 56 Chapter 10: Elasticity & Oscillations • • • • • • • • • • Elastic deformations of solids Hooke's law for tensile and compressive forces Beyond Hooke's law Shear and volume deformations Simple harmonic motion The period and frequency for SHM Graphical analysis of SHM The pendulum Damped oscillations Forced oscillations and resonance MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 57 Extra MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 58 Aeroelasticity Aeroelasticity is the science which studies the interactions among inertial, elastic, and aerodynamic forces. It was defined by Arthur Collar in 1947 as "the study of the mutual interaction that takes place within the triangle of the inertial, elastic, and aerodynamic forces acting on structural members exposed to an airstream, and the influence of this study on design." MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 59 Aeroelastic Flutter Flutter Flutter is a self-feeding and potentially destructive vibration where aerodynamic forces on an object couple with a structure's natural mode of vibration to produce rapid periodic motion. Flutter can occur in any object within a strong fluid flow, under the conditions that a positive feedback occurs between the structure's natural vibration and the aerodynamic forces. That is, that the vibrational movement of the object increases an aerodynamic load which in turn drives the object to move further. If the energy during the period of aerodynamic excitation is larger than the natural damping of the system, the level of vibration will increase, resulting in self-exciting oscillation. The vibration levels can thus build up and are only limited when the aerodynamic or mechanical damping of the object match the energy input, this often results in large amplitudes and can lead to rapid failure. Because of this, structures exposed to aerodynamic forces - including wings, aerofoils, but also chimneys and bridges - are designed carefully within known parameters to avoid flutter. It is however not always a destructive force; recent progress has been made in small scale (table top) wind generators for underserved communities in developing countries, designed specifically to take advantage of this effect. In complex structures where both the aerodynamics and the mechanical properties of the structure are not fully understood flutter can only be discounted through detailed testing. Even changing the mass distribution of an aircraft or the stiffness of one component can induce flutter in an apparently unrelated aerodynamic component. At its mildest this can appear as a "buzz" in the aircraft structure, but at its most violent it can develop uncontrollably with great speed and cause serious damage to or the destruction of the aircraft. In some cases, automatic control systems have been demonstrated to help prevent or limit flutter related structural vibration. Flutter can also occur on structures other than aircraft. One famous example of flutter phenomena is the collapse of the original Tacoma Narrows Bridge. MFMcGraw-PHY1401 Chap 10d - Elas & Vibrations - Revised 7-12-10 60