1 PHYS1002 Physics 1 (Fundamentals) Oscillations and Waves Semester 1, 2011 Ian Cooper Textbook: College Physics (Knight, Jones, Field) Chapters oscillations_01

2 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 OVERVIEW Mindmaps – A3 summaries

3 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 Equation Mindmaps

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 When force F applied, wire (spring) extends a distance x displacement x y, s,  x … Elastic behaviour: Wire returns to original length when force is removed Plastic behaviour: Distortion remains when force is removed restoring force F e applied force F equilibrium position: x = 0 + x equilibrium length CP

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] spring or elastic constant [N.m -1 ] linearly elastic region as F is increased beyond the elastic limit the extension will become permanent CP applied force [N] (0,0) plastic region spring does not recover F e  (F sp ) x elastic limit

11 F F x x F = k x F e = - k x equilibriumcompressedExtended (stretched) Hooke’s Law FeFe FeFe CP

12 O 0 FF xx k = slope of F vs x graph k1k1 k2k2 k3k3 k 1 > k 2 > k 3 Stiffest or most rigid spring rise run “pliant” materials: large deformation – small forces CP

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

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 U e 2 Reference point U e = 0 when x = 0

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

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

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

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

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

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 m and held with a force of 450 N. What is the elastic constant k of the bow? use the ISEE method

Solution 3 Identify / Setup Bow and arrow k = ? N.m -1 F = 450 N x = m Hooke’s Law F = k x Execute k = F / x = 450 / 0.7 N.m -1 k = 6.43  10 2 N.m -1 Evaluate OK 21

22 Problem 4 During the filming of a movie a 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

Solution 4 Identify / Setup m = 100 kg L = 50.0 m k = 1000 N.cm -1 1 cm = m 1 cm -1 = 10 2 m -1 k = 10 5 N.m -1 g = 9.80 m.s -2 x max = ? m 1 2 L x max K 1 = 0 U G1 = m g (L + x max ) U e1 = 0 K 2 = 0 U G2 = 0 U e2 = ½ k x max 2 K = ½ m v 2 U G = m g h U e = ½ k x 2 Conservation of energy E = U + K = constant 23

Execute Evaluate OK energy conserved E 1 = E 2  m g (L + x max ) = ½ k x max 2 ½ k x max 2 – m g x max – m g L =0 Quadratic equation a x 2 + b x + c = 0 x = {-b   (b 2 – 4 a c)} / (2a) a = k /2 b = - m g c = - m g L a = 5  10 4 b = c = -4.9  10 4 x max = 1.00 m or x max = m max stretch x max = 1.00 m 24

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. (d)What might happen to this person? (e)How fast does the person hit the ground? use the ISEE method

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