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Chapter 10 Elasticity & Oscillations

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. Elastic Deformation of Solids.

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Chapter 10 Elasticity & Oscillations

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  1. Chapter 10 Elasticity & Oscillations

  2. 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 Chap 10d - Elas & Vibrations - Revised 7-12-10

  3. 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. Chap 10d - Elas & Vibrations - Revised 7-12-10

  4. 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. Chap 10d - Elas & Vibrations - Revised 7-12-10

  5. Stress and Strain Define: The fractional change in length Force per unit cross-sectional area Stretching ==> Tensile Stress Squeezing ==> Compressive Stress Chap 10d - Elas & Vibrations - Revised 7-12-10

  6. Hooke’s Law Hooke’s Law (Fx) can be written in terms of stress and strain (stress  strain). The spring constant k is now 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. Chap 10d - Elas & Vibrations - Revised 7-12-10

  7. 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 5.8104 N and the length of the beam is 2.5 m, and the cross-sectional area of the beam is 7.5103 m2. Find the vertical compression of the beam. Force of ceiling on beam Force of floor on beam For steel Y = 200109 Pa. Chap 10d - Elas & Vibrations - Revised 7-12-10

  8. Tensile Stress Example (text problem 10.7): A 0.50 m long guitar string, of cross-sectional area 1.0106 m2, has a Young’s modulus of 2.0109 Pa. By how much must you stretch a guitar string to obtain a tension of 20 N? Chap 10d - Elas & Vibrations - Revised 7-12-10

  9. 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. Chap 10d - Elas & Vibrations - Revised 7-12-10

  10. Chap 10d - Elas & Vibrations - Revised 7-12-10

  11. 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 Chap 10d - Elas & Vibrations - Revised 7-12-10

  12. 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 2.5108 Pa. What is the minimum diameter the wire should have to support her? Want Chap 10d - Elas & Vibrations - Revised 7-12-10

  13. Different Representations Original equation Fractional change Change in length New length Chap 10d - Elas & Vibrations - Revised 7-12-10

  14. Shear Deformations A shear deformation occurs when two forces are applied on opposite surfaces of an object. Chap 10d - Elas & Vibrations - Revised 7-12-10

  15. Stress and Strain Define: Hooke’s law (stressstrain) for shear deformations is where S is the shear modulus Chap 10d - Elas & Vibrations - Revised 7-12-10

  16. F F 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? From Hooke’s Law: Chap 10d - Elas & Vibrations - Revised 7-12-10

  17. Volume Deformations An object completely submerged in a fluid will be squeezed on all sides. The result is a volume strain; Chap 10d - Elas & Vibrations - Revised 7-12-10

  18. Volume Deformations For a volume deformation, Hooke’s Law is (stressstrain): where B is called the bulk modulus. The bulk modulus is a measure of how easy a material is to compress. Chap 10d - Elas & Vibrations - Revised 7-12-10

  19. An anchor, made of cast iron of bulk modulus 60.0109 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 1.75106 Pa. Find the change in the volume of the anchor. Chap 10d - Elas & Vibrations - Revised 7-12-10

  20. Examples • I-beam • Arch - Keystone • Flying Buttress Chap 10d - Elas & Vibrations - Revised 7-12-10

  21. 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 proportionality Young’s modulus (Y) Shear modulus (S) Bulk Modulus (B) Deformations Summary Table Chap 10d - Elas & Vibrations - Revised 7-12-10

  22. 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. Chap 10d - Elas & Vibrations - Revised 7-12-10

  23. 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 Chap 10d - Elas & Vibrations - Revised 7-12-10

  24. Equilibrium position y x x The Spring The motion of a mass on a spring is an example of SHM. The restoring force is F = kx. Chap 10d - Elas & Vibrations - Revised 7-12-10

  25. Equation of Motion & Energy Assuming the table is frictionless: Classic form for SHM Also, Chap 10d - Elas & Vibrations - Revised 7-12-10

  26. Spring Potential Energy Chap 10d - Elas & Vibrations - Revised 7-12-10

  27. 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. Chap 10d - Elas & Vibrations - Revised 7-12-10

  28. Chap 10d - Elas & Vibrations - Revised 7-12-10

  29. 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. Chap 10d - Elas & Vibrations - Revised 7-12-10

  30. Spring Compensates for Gravity Chap 10d - Elas & Vibrations - Revised 7-12-10

  31. Representing Simple Harmonic Motion Chap 10d - Elas & Vibrations - Revised 7-12-10

  32. A simple harmonic oscillator can be described mathematically by: where A is the amplitude of the motion, the maximum displacement from equilibrium, A = vmax, and A2 = amax. Or by: Chap 10d - Elas & Vibrations - Revised 7-12-10

  33. Linear Motion - Circular Functions Chap 10d - Elas & Vibrations - Revised 7-12-10

  34. Projection of Circular Motion Chap 10d - Elas & Vibrations - Revised 7-12-10

  35. 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. Chap 10d - Elas & Vibrations - Revised 7-12-10

  36. 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: Since E = constant, at equilibrium (x = 0) the KE must be a maximum. Here v = vmax = A. Chap 10d - Elas & Vibrations - Revised 7-12-10

  37. Example continued: The amplitude A is given, but  is not. Chap 10d - Elas & Vibrations - Revised 7-12-10

  38. 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 1.8104 m at that frequency. (a) What is the maximum force acting on the diaphragm? The value is Fmax=1400 N. Chap 10d - Elas & Vibrations - Revised 7-12-10

  39. Example continued: (b) What is the mechanical energy of the diaphragm? Since mechanical energy is conserved, E = Kmax = Umax. The value of k is unknown so use Kmax. The value is Kmax= 0.13 J. Chap 10d - Elas & Vibrations - Revised 7-12-10

  40. Example (text problem 10.47): The displacement of an object in SHM is given by: What is the frequency of the oscillations? Comparing to y(t) = A sint gives A = 8.00 cm and  = 1.57 rads/sec. The frequency is: Chap 10d - Elas & Vibrations - Revised 7-12-10

  41. Example continued: Other quantities can also be determined: The period of the motion is Chap 10d - Elas & Vibrations - Revised 7-12-10

  42. 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. Chap 10d - Elas & Vibrations - Revised 7-12-10

  43. 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 φ-coordinate is perpendicualr to the rod and is positive in the counterclock wise direction. Chap 10d - Elas & Vibrations - Revised 7-12-10

  44. Apply Newton’s 2nd Law to the pendulum bob. If we assume that φ <<1 rad, then sin φ  φ and cos φ 1, the angular frequency of oscillations is then: The period of oscillations is Chap 10d - Elas & Vibrations - Revised 7-12-10

  45. 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? Solving for L: Chap 10d - Elas & Vibrations - Revised 7-12-10

  46. Lcos L L y=0 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). Chap 10d - Elas & Vibrations - Revised 7-12-10

  47. 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. Chap 10d - Elas & Vibrations - Revised 7-12-10

  48. The Physical Pendulum http://hyperphysics.phy-astr.gsu.edu/HBASE/pendp.html Chap 10d - Elas & Vibrations - Revised 7-12-10

  49. Damped Oscillations When dissipative forces such as friction are not negligible, the amplitude of oscillations will decrease with time. The oscillations are damped. Chap 10d - Elas & Vibrations - Revised 7-12-10

  50. Graphical representations of damped oscillations: Chap 10d - Elas & Vibrations - Revised 7-12-10

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