Elasticity and Oscillations

1. Elasticity and Oscillations

2. Intrinsic Strength of Materials • Hanging by a hair • Can a single hair support your weight? Steel wire? A hair, cable etc. can hold more weight if it has greater cross section.

3. Intrinsic Strength of Materials • Would we build a suspension bridge from Rapunzal’s hair? Steel cable is intrinsically stronger than hair, rope

4. Intrinsic Strength of Materials • Compare strength of materials: • consider not just how much weight they can support • but how much weight for a given cross section F/A. Stress = F/A

5. Hooke’s LawF = - k x F F Apply a force to both ends of a long wire. These forces will stretch the wire from length L to L+L. F = kΔL

6. Strain If we pull on a spring it increases in length If we pull on a wire increases in length. Longer wires (and springs) will stretch more than short ones

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

8. Hooke’s Law (Fx) can be written in terms of stress and strain (stress  strain). The spring constant k is now (F=kx) 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.

9. 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?

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

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

12. Simple Harmonic Motion • What is motion of the cart if we move it slightly? • Stable repeats, oscillates • Gone! • How do we describe a)?

14. Oscillations • Time it takes for the ball to roll down, up opposite side, down and back is the period T • Tis the time to complete one full cycle. • Equivalent information is • number of cycles in one second ,frequency f. • e.g. T= 1/10 f = 10; T = 5 f= 1/5 f=1/T T = 1/f Frequency is measured in Hertz = 1 cycle /second

15. We can plot the position of the ball as a function of time. A is the distance from the bottom of the cup.

16. 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.

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

18. Assuming the table is frictionless: Also,

19. 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.

20. 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.

21. SHM graphically

22. t=0: m starts with x=A, v=0 and a negative. As soon as m released v becomes negative. t=1/4 T: equilibrium x=0, v max(-), a=0 t= ½ T: x=-A, v=0 a positive, begins + motion t= ¾ T: x=0 equilibrium v max (+), a=0 t= T; back at the beginning. Here we go again.

23. Simple Harmonic Motion • The x graph looks like a cosine • x= A cos(2πt/T) • A is maximum displacement, ω =2π/T • The v graph also is SHO • v = vmax sinωt • vmax should get larger if A is large (ω fixed) • vmax should get larger if ω is large (A fixed) • Vmax ~ωA

24. At endpoints v=0, x max At equilibrium x=0 v max v2max =k/m A2 v= √k/m A a =-ω2 x

25. 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:

26. 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.

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: Since E = constant, at equilibrium (x = 0) the KE must be a maximum. Here v = vmax = A.

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

29. Question A mass on a spring has amplitude A. If A is doubled, the total energy of the system is: A) doubled. B) quadrupled. C) the same. D) halved. E) 1/4 as much.

30. 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.

31. (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.

32. 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:

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