Chapter 10 Lecture Outline. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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1 Chapter 10 Lecture Outline Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1

2 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

3 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

4 10. 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

5 Define: strain L L The fractional change in length stress F A Force per unit crosssectional area 5

6 Hooke s Law (Fx) can be written in terms of stress and strain (stress strain). F A Y L L The spring constant k is now k YA 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

7 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 N and the length of the beam is.5 m, and the cross-sectional area of the beam is m. Find the vertical compression of the beam. Force of ceiling on beam F A L Y L L F A L Y Force of floor on beam For steel Y = Pa. L F A L Y N m.5 m N/m m 7

8 Example (text problem 10.7): A 0.50 m long guitar string, of cross-sectional area m, has a Young s modulus of Pa. By how much must you stretch a guitar string to obtain a tension of 0 N? F A L L Y L F A L Y N m m 5.0 mm 0.5 m N/m 8

9 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 crosssectional area is called tensile strength. 9

10 The ultimate strength of a material is the maximum stress that it can withstand before breaking. 10

11 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 Pa. What is the minimum diameter the wire should have to support her? Want F stress elastic A limit A F elastic limit mg elastic limit D mg elastic limit D 4mg elastic limit m 1.7 mm 11

12 10.4 Shear and Volume Deformations A shear deformation occurs when two forces are applied on opposite surfaces of an object. 1

13 Shear Stress Shear Force Surface Area F A Define: Shear Strain displacement of surfaces separation of surfaces x L Hooke s law (stressstrain) for shear deformations is F A S x L where S is the shear modulus 13

14 Example (text problem 10.5): 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 A S x L F From Hooke s Law: F x SA L 0.64 cm 5.0 cm 940 N/m m 0.30 N 14

15 An object completely submerged in a fluid will be squeezed on all sides. volume stress pressure F A The result is a volume strain; volume strain V V 15

16 For a volume deformation, Hooke s Law is (stressstrain): P B V V where B is called the bulk modulus. The bulk modulus is a measure of how easy a material is to compress. 16

17 Example (text problem 10.4): An anchor, made of cast iron of bulk modulus Pa and a volume of 0.30 m 3, is lowered over the side of a ship to the bottom of the harbor where the pressure is greater than sea level pressure by Pa. Find the change in the volume of the anchor. P V V B V VP B m Pa m Pa 17

18 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 proportionality Young s modulus (Y) Shear modulus (S) Bulk Modulus (B) 18

19 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

20 The motion of a mass on a spring is an example of SHM. Equilibrium position y x x The restoring force is F = kx. 0

21 Assuming the table is frictionless: E t K a t x F t x U kx k m xt ma Also, t 1 mv t x 1 kx t 1

22 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.

23 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. 3

24 SHM graphically 4

25 A simple harmonic oscillator can be described mathematically by: Or by: x v a x v a t t t t t t Acost x t v t A sin t A cost Asin t x A cost t v A sin t t where A is the amplitude of the motion, the maximum displacement from equilibrium, A = v max, and A = a max. 5

26 The period of oscillation is T. where is the angular frequency of the oscillations, k is the spring constant and m is the mass of the block. k m 6

27 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: E K U 1 mv 1 kx 1 mv Since E = constant, at equilibrium (x = 0) the KE must be a maximum. Here v = v max = A. 7

28 Example continued: The amplitude A is given, but is not. T s 1. 6 rads/sec and v Aω 5. 0 cm 1. 6 rads/sec 6. 8 cm/sec 8

29 Example (text problem 10.41): The diaphragm of a speaker has a mass of 50.0 g and responds to a signal of.0 khz by moving back and forth with an amplitude of m at that frequency. (a) What is the maximum force acting on the diaphragm? F F ma f maf max mamax m A 4 The value is F max =1400 N. 9

30 Example continued: (b) What is the mechanical energy of the diaphragm? Since mechanical energy is conserved, E = K max = U max. U K max max 1 ka The value of k is unknown so use K max. 1 mv max K 1 1 A ma f max mvmax m 1 The value is K max = 0.13 J. 30

31 Example (text problem 10.47): The displacement of an object in SHM is given by: y t 8.00 cmsin 1.57 rads/sec What is the frequency of the oscillations? t Comparing to y(t) = A sint gives A = 8.00 cm and = 1.57 rads/sec. The frequency is: f 1.57 rads/sec 0.50 Hz 31

32 Example continued: Other quantities can also be determined: The period of the motion is T 1.57 rads/sec 4.00 sec x v a max max max A A A cm cm157. rads/sec 1. 6 cm/sec cm157. rads/sec cm/sec 3

33 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

34 A simple pendulum: L An FBD for the pendulum bob: y T m w x 34

35 Apply Newton s nd Law to the pendulum bob. F F x y mgsin ma T mg cos m t v r If we assume that <<1 rad, then sin and cos 1, the angular frequency of oscillations is then: g L The period of oscillations is T L g 35

36 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? T L g Solving for L: L gt m/s 10. s m 36

37 Example (text problem 10.94): 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). L L Lcos y=0 y L( 1cos ) 37

38 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. 38

39 10.9 Damped Oscillations When dissipative forces such as friction are not negligible, the amplitude of oscillations will decrease with time. The oscillations are damped. 39

40 Graphical representations of damped oscillations: 40

41 10.10 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. 41

42 Summary Stress and Strain Hooke s Law Simple Harmonic Motion SHM Examples: Mass-Spring System, Simple Pendulum and Physical Pendulum Energy Conservation Applied to SHM 4

Chapter 13 Oscillations about Equilibrium. Copyright 2010 Pearson Education, Inc.

Chapter 13 Oscillations about Equilibrium Periodic Motion Units of Chapter 13 Simple Harmonic Motion Connections between Uniform Circular Motion and Simple Harmonic Motion The Period of a Mass on a Spring