Chapter 10: Elasticity and Oscillations

1. 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: The fractional change in length Force per unit cross-sectional area

5. 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.

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 5.8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.

7. Example (text problem 10.6): 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 20.0 N?

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 2.5108 Pa. What is the minimum diameter the wire should have to support her? Want

11. §10.4 Shear and Volume Deformations A shear deformation occurs when two forces are applied on opposite surfaces of an object.

12. Define: Hooke’s law (stressstrain) for shear deformations is where S is the shear modulus

13. 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. If the shear modulus of the gelatin is 940 Pa, what is the magnitude of the tangential force? From Hooke’s Law:

14. An object completely submerged in a fluid will be squeezed on all sides. The result is a volume strain;

15. 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.

16. Example (text problem 10.24): 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.

17. Deformations summary table

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

20. Assuming the table is frictionless: Also,

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: where A is the amplitude of the motion, the maximum displacement from equilibrium, A=vmax, and A2 =amax. Or by:

25. 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.

26. Example (text problem 10.28): 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.

27. Example continued: The amplitude A is given, but  is not.

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 1.810-4 m at that frequency. (a) What is the maximum force acting on the diaphragm? The value is Fmax=1400 N.

29. Example continued: (b) What is the mechanical energy of the diaphragm? Since mechanical energy is conserved, E = KEmax = Umax. The value of k is unknown so use KEmax. The value is KEmax= 0.13 J.

30. 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:

31. Example continued: Other quantities can also be determined: The period of the motion is

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. y  T L  m x w A simple pendulum: An FBD for the pendulum bob: Assume <<1 radian

34. Apply Newton’s 2nd Law to the pendulum bob. If we assume that <<1 rad, then sin   and cos 1 then the angular frequency of oscillations is found to be: The period of oscillations is

35. 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:

36.  Lcos L L y=0 Example (text problem 10.84): 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).

37. 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. §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. Graphical representations of damped oscillations:

40. §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. 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