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Chapter 10: Elasticity and Oscillations

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Chapter 10: Elasticity and Oscillations Chapter 10 Lecture Outline 1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 10: 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 2 §10.1 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. 3 §10.2 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. 4 Define: L The fractional strain L change in length F Force per unit cross- stress A sectional area 5 Hooke’s Law (Fx) can be written in terms of stress and strain (stress strain). F L Y A L YA The spring constant k is now 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. 6 Example (text problem 10.1): 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 F L ceiling Y on beam A L F L L A Y Force of 9 floor on For steel Y = 20010 Pa. beam F L 5.8104 N 2.5 m L 1.0104 m A Y 7.5103 m2 200109 N/m2 7 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 F L 20.0 N 0.5 m L A Y 1.0106 m2 2.0109 N/m2 5.0103 m 5.0 mm 8 §10.3 Beyond Hooke’s Law If the stress on an object exceeds the elastic limit, then the object will not return to its original length. 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. 9 The ultimate strength of a material is the maximum stress that it can withstand before breaking. 10 Example (text problem 10.10): 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 Want stress elastic limit A F mg A elastic limit elastic limit 2 D mg 2 elastic limit 4mg D 1.7103 m 1.7 mm elastic limit 11 §10.4 Shear and Volume Deformations A shear deformation occurs when two forces are applied on opposite surfaces of an object. 12 Shear Force F Shear Stress Surface Area A Define: displacement of surfaces x Shear Strain separation of surfaces L Hooke’s law (stressstrain) for shear deformations is F x S where S is the A L shear modulus 13 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. If the shear modulus of the gelatin is 940 Pa, what is the magnitude of the tangential force? F F From Hooke’s Law: x F x F SA S L A L 0.64 cm 940 N/m2 0.0025 m2 0.30 N 5.0 cm 14 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 15 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. 16 Example (text problem 10.24): 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 m3 1.75106 Pa V B 60.0109 Pa 6.71106 m3 17 Deformations summary table Tensile or compressive Shear Volume Stress Force per unit Shear force divided Pressure cross-sectional by the area of the area surface on which it acts Strain Fractional Ratio of the relative Fractional change in displacement to the change in length separation of the two volume parallel surfaces Constant of Young’s Shear modulus (S) Bulk proportionality modulus (Y) Modulus (B) 18 §10.5 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. 19 The motion of a mass on a spring is an example of SHM. Equilibrium position y x x The restoring force is F = kx. 20 Assuming the table is frictionless: F kx ma x x k a t xt x m 1 1 Also, Et KtUt mvt2 kxt2 2 2 21 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. 22 §10.6-7 Representing Simple Harmonic Motion When a mass-spring system is oriented vertically, it will exhibit SHM with the same period and frequency as a horizontally placed system. 23 SHM graphically 24 A simple harmonic oscillator can be described mathematically by: xt Acost x vt A sint t v at A 2 cost t where A is the amplitude of the motion, the maximum Or by: displacement from equilibrium, A = v , and A2 = a xt Asint max max. x vt A cost t v at A 2 sint t 25 2 The period of oscillation is T . where is the angular frequency of the k oscillations, k is the spring constant and m m is the mass of the block. 26 Example (text problem 10.30): 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 1 1 E K U mv2 kx2 mv2 2 2 2 Since E = constant, at equilibrium (x = 0) the KE must be a maximum. Here v = vmax = A. 27 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 cm12.6 rads/sec 62.8 cm/sec 28 Example (text problem 10.41): 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 2 2 F Fmax mamax mA mA2f 4 mAf The value is Fmax=1400 N. 29 Example continued: (b) What is the mechanical energy of the diaphragm? Since mechanical energy is conserved, E = Kmax = Umax. 1 U kA2 The value of k is unknown so use K . max 2 max 1 2 Kmax mvmax 1 1 1 2 K mv2 mA2 mA2 2f 2 max 2 max 2 2 The value is Kmax= 0.13 J. 30 Example (text problem 10.47): The displacement of an object in SHM is given by: yt 8.00 cmsin1.57 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 31 Example continued: Other quantities can also be determined: 2 2 The period of the motion is T 4.00 sec 1.57 rads/sec xmax A 8.00 cm vmax A 8.00 cm1.57 rads/sec 12.6 cm/sec 2 2 2 amax A 8.00 cm1.57 rads/sec 19.7 cm/sec 32 §10.8 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. 33 A simple pendulum: An FBD for the pendulum bob: L y T m x w 34 Fx mgsin mat Apply Newton’s 2nd Law to the pendulum bob. v2 F T mgcos m y r If we assume that <<1 rad, then sin and cos 1, the angular frequency of oscillations is then: g L 2 L The period of oscillations is T 2 g 35 Example (text problem 10.60): A clock has a pendulum that performs one full swing every 1.0 sec.
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