PHYS1002 Physics 1 (Fundamentals) Oscillations and Waves Semester 1, 2011

1. oscillations_01 PHYS1002 Physics 1 (Fundamentals)Oscillations and WavesSemester 1, 2011 Ian Cooper cooper@physics.usyd.edu.au Textbook: College Physics (Knight, Jones, Field) Chapters 8.310.4 14 15 16 17.1 17.4

2. OVERVIEW Mindmaps – A3 summaries • Oscillations • Elastic materials & Hooke’s law • Simple harmonic motion • Damped Oscillations • Resonance • Waves • Transverse & longitudinal waves • Behaviour of waves • Sound • Superposition principle and interference • Standing Waves • Beats and Doppler effect • Electromagnetic spectrum • Refractive index • Thin film interference

3. Equation Mindmaps For each equations on the Exam Formula Sheet – you should construct an Equation Mindmap Symbols – interpretation, units, signs Visualization & interpretation Assumptions Special constants Graphical interpretation Applications, Comments Numerical Examples

4. What happens to our skin when we become old ? What Physics is in the pictures?

5. What is are the basic components of a car suspension system ? How is the elastic potential energy related to earthquake damage? Queenstown NZ – World Home of Bungy What is the physics?

6. Why measure the restoring force of a DNA molecule ? What does Hooke’s Law have to do with a nasal strip (device for improving air flow through nasal passages)?

7. oscillations_01: MINDMAP SUMMARY Elastic, Plastic, Reference frame (coordinate system, origin, equilibrium position), displacement (extension, compression, applied force, restoring force, gravitational force, net (resultant) force, Newton’s Second Law), Hooke’s Law, spring constant (spring stiffness), equilibrium, velocity, acceleration, work, kinetic energy, potential energy (reference point), gravitational potential energy, elastic potential energy, total energy, conservation of energy, ISEE, solve quadratic equations

8. To study any physical phenomena, a physicist starts with a simple model. For a vibrating mechanical system, we consider the simplest model in which an object disturbed from its equilibrium position is acted upon by a restoring force that acts to return the object back to its equilibrium position. The restoring force is proportional to the displacement of the object from equilibrium and acts in the opposite direction to the displacement. This restoring force obeys Hooke’s Law. The motion of the object acted on by this type of restoring force is periodic and is called Simple Harmonic Motion. 1

9. Elastic and plastic behaviour equilibrium position: x = 0 equilibrium length + x restoring force Fe applied force F displacement xy, s, x… When force F applied, wire (spring) extends a distance x Elastic behaviour: Wire returns to original length when force is removed Plastic behaviour: Distortion remains when force is removed CP239-245

10. Hooke’s Law (simplest model for restoring force) Extension or compression is proportional to restoring & applied forces extension or compression [m] restoring force [N] applied force [N] spring or elastic constant [N.m-1] as F is increased beyond the elastic limit the extension will become permanent linearly elastic region plastic region spring does not recover elastic limit Fe (Fsp)x (0,0) CP239-245

11. equilibrium compressed Extended (stretched) Fe x F Fe x F Hooke’s Law F = k x Fe = - k x CP239-245

12. Stiffest or most rigid spring F F k1 k2 rise “pliant”materials: large deformation – small forces run k3 0 x x O k1 > k2 > k3 k = slope of F vs x graph CP239-245

13. Elastic potential energy work done in extending wire = area under curve [J  N.m] = potential energy Ue stored in extended wire [J] linearly elastic region Examples: Pogo stick, longbow, crossbow, pole vaulting, ….. CP303-304

14. Work done W by an applied force F in extending a spring through a displacement x The work done W increases the potential energy of the mass/spring system Ue Reference point Ue = 0 when x = 0 2

15. Problem solving strategy: I S E E Identity: What is the question asking (target variables) ? What type of problem, relevant concepts, approach ? Set up: Diagrams Equations Data (units) Physical principals Execute: Answer question Rearrange equations then substitute numbers Evaluate: Check your answer – look at limiting cases sensible ? units ? significant figures ? PRACTICE ONLY MAKES PERMANENT

16. DNA is a long-chain molecule that is normally tightly coiled. Amazingly it is possible to grab the two ends of a DNA molecule and gently stretch it while measuring the restoring force using optical tweezers. Knowing the restoring force tells how various enzymes act to cut and then reseal coils in the DNA structure. Problem 1 A DNA molecule is anchored at one end, then a force of 1.55 nN pulls on the other end, causing the molecule to stretch by 5.2 nm. What is the spring constant of the DNA molecule ? use the ISEE method

17. Solution 1 DNA molecule Identify / Setup F = 1.55 nN = 1.5510-9 N k = ? N.m-1 x = 5.2 nm = 5.210-9 m Hooke’s Law F = k x Execute k = F / x = (1.5510-9 / 5.210-9 ) N.m-1 k = 0.30 N.m-1 Evaluate OK

18. Problem 2 A nasal strip can improve the air flow through nasal passages. The nasal strip consists of two flat polyester springs enclosed by an adhesive tape covering. Measurements show that a nasal strip can exert a force of 0.25 N on the nose, causing it to expand by 3.7 mm. Calculate the effective force constant of the nasal strip and the force required to expand the nose by 4.2 mm. use the ISEE method

19. Solution 2 nasal strip Identify / Setup F = 0.25 N k = ? N.m-1 x = 3.7 mm = 3.710-3 m x = 4.2 mm = 4.210-3 m F = ? N Hooke’s Law F = k x Execute k = F / x = (0.25 / 3.710-3 ) N.m-1 =68 N.m-1 F = k x = (68)(4.210-3) N = 0.28 N Evaluate OK

20. Problem 3 When a bowstring is pulled back in preparation for shooting an arrow, the system behaves in a Hookean fashion. Suppose the string is drawn 0.700 m and held with a force of 450 N. What is the elastic constant k of the bow? use the ISEE method

21. Solution 3 Bow and arrow Identify / Setup F = 450 N k = ? N.m-1 x = 0.700 m Hooke’s Law F = k x Execute k = F / x = 450 / 0.7 N.m-1 k = 6.43102 N.m-1 Evaluate OK

22. Problem 4 During the filming of a movie a 100.0 kg stuntman steps off the roof of a building and free-falls. He is attached to a safety line 50.0 m long that has an elastic constant 1000 N.cm-1. What is the maximum stretch of the line at the instant he comes to rest, assuming Hooke’s Law is valid? Hint: Consider how the various kinds of energy change use the ISEE method

23. Solution 4 1 K1 = 0 UG1 = m g (L + xmax) Ue1 = 0 Identify / Setup L m = 100 kg L = 50.0 m k = 1000 N.cm-1 1 cm = 10-2 m 1 cm-1 = 102 m-1 k = 105 N.m-1 g = 9.80 m.s-2 xmax = ? m xmax K2 = 0 UG2 = 0 Ue2 = ½ k xmax2 2 K = ½ m v2UG = m g hUe = ½ k x2 Conservation of energy E = U + K = constant

24. Execute energy conserved E1 = E2 m g (L + xmax) = ½ k xmax2 ½ k xmax2 – m g xmax – m g L =0 Quadratic equation a x2 + b x + c = 0 x = {-b  (b2 – 4 a c)} / (2a) a = k /2 b = - m gc = - m g L a = 5104b = - 980 c = -4.9104 xmax = 1.00 m or xmax = - 0.98 m max stretch xmax = 1.00 m Evaluate OK

25. Problem 5 • Consider a person taking a bungee jump. The mass of the jumper is 60.0 kg. The natural length of the bungee cord is 9.00 m. At the bottom of the jump, the bungee cord has extended by 18.0 m. • (a) What is the spring constant? • (b) What is the maximum elastic force restoring force • exerted on the jumper? • (c) What is the acceleration experienced by the • jumper at the bottom of the jump? • The person misses the ground by 3.00 m. Another person who has a mass of 120 kg takes the same cord (without permission) and takes the plunge. • What might happen to this person? • How fast does the person hit the ground? use the ISEE method

26. Answers to bungee jump Problem 5 • (a) k = 98 N.m-1 • (b) Femax = 1764 N • (c) a = 19.6 m.s-2 = 2g • (d) hdrop = 40 m > 30 m • (e) v = 15 m.s-1 = 54 km.h-1 • jumper could be killed or seriously injured