Lecture 4

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Lecture 4

  1. 1. 1 LECTURE 4 CP Ch 14 Damped Oscillations Forced Oscillations and Resonance
  2. 2. 2 ( ) ( ) ( ) ( ) ( ) ( ) max ma 2 2 2 2 max 2 2 2 2 2 2 max x a max 2 m x cos sin cos 1 1 cos 2 2 1 1 1 sin sin 2 2 2 x t v t a t a x PE k x k x t KE mv m x t x x x k x t ω ω ω ω ω ω ω ω ω ω = = − = − = − = = = = = 2 2 max max 1 1 = constant 2 2 totalE KE PE k x mv= + = = 2 2 2 2 2 max max 1 1 1 2 2 2 mv k x k x v x xω+ = ⇒ = ± − 2 2 2 f T k m π ω π ω = = = CP 445 SHM
  3. 3. 3 0 10 20 30 40 50 60 70 80 90 100 -10 0 10 SHM positionx 0 10 20 30 40 50 60 70 80 90 100 -5 0 5 velocityv 0 10 20 30 40 50 60 70 80 90 100 -1 0 1 accelerationa time t CP445
  4. 4. 4 0 2 4 6 8 0 0.02 0.04 0.06 0.08 0.1 0.12 b = 0energyKUE(J) time t (s) KE PE E CP 455
  5. 5. 5 Mathematical modelling for harmonic motion Newton’s Second Law can be applied to the oscillating system Σ F = restoring force + damping force + driving force Σ F(t) = - k x(t) - b v(t) + Fd(t) For a harmonic driving force at a single frequency Fd(t) = Fmaxcos(ωt + ε). This differential equation can be solved to give x(t), v(t) and a(t). CP 463
  6. 6. 6 Damped oscillations Oscillations in real systems die away (the amplitude steadily decreases) over time - the oscillations are said to be damped For example: The amplitude of a pendulum will decrease over time due to air resistance If the oscillating object was in water, the greater resistance would mean the oscillations damp much quicker. CP 463
  7. 7. 7 Damped oscillations CP 463
  8. 8. 8 0 2 4 6 8 0 0.02 0.04 0.06 0.08 0.1 0.12 b = 6 energyKUE(J) time t (s) KE PE E CP 463
  9. 9. 9 Forced oscillations Driven Oscillations & Resonance If we displace a mass suspended by a spring from equilibrium and let it go it oscillates at its natural frequency f = 1 2π k m If a periodic force at another frequency is applied, the oscillation will be forced to occur at the applied frequency - forced oscillations CP 465
  10. 10. 10 Resonance Forced oscillations are small unless the driving frequency is close to the natural frequency When the driving frequency is equal to the natural frequency the oscillations can be large - this is called resonance Away from resonance, energy transfer to the oscillations is inefficient. At resonance there is efficient transfer which can cause the oscillating system to fail - see wine glass experiment. Famous example of resonance: soldiers marching on bridge CP 465
  11. 11. 11 Resonance phenomena occur widely in natural and in technological applications: Emission & absorption of light Lasers Tuning of radio and television sets Mobile phones Microwave communications Machine, building and bridge design Musical instruments Medicine – nuclear magnetic resonance magnetic resonance imaging – x-rays – hearing Nuclear magnetic resonance scan CP 465
  12. 12. 12 0 0.5 1 1.5 2 2.5 3 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 amplitudeA(m) ωd /ωo b = 2 b = 8 b = 10 CP 465 Response Curve fd / fO
  13. 13. 13 Sinusoidal driving force fd / fo = 0.1 0 20 40 60 80 100 -1 -0.5 0 0.5 1 b = 2 positionx(m) time t (s) Sinusoidal driving force fd / fo = 1 0 20 40 60 80 100 -1 -0.5 0 0.5 1 b = 2 positionx(m) time t (s) CP 465
  14. 14. 14 Sinusoidal driving force fd / fo = 2 0 20 40 60 80 100 -1 -0.5 0 0.5 1 b = 2 positionx(m) time t (s) Impulsive force – constant force applied for a short time interval. 0 20 40 60 80 100 -1 -0.5 0 0.5 1 b = 2 positionx(m) time t (s) CP 465
  15. 15. 15 http://www.acoustics.salford.ac.uk/feschools/waves/wine3video.htm http://www.acoustics.salford.ac.uk/feschools/waves/shm3.htm An optical technique called interferometry reveals the oscillations of a wine glass Great Links to visit
  16. 16. 16 Eardrum Auditory canal Cochlea Basilar membrane vibrations of small bones of the middle ear vibration of eardrum due to sound waves Inner air – basilar membrane – as the distance increases from the staples, membrane becomes wider and less stiff – resonance frequency of sensitive hair cells on membrane decreases staples 1 2 o k f mπ = 3000 Hz 30 Hz Resonance and Hearing CP467 staples
  17. 17. 17 Self excited oscillations Sometimes apparently steady forces can cause large oscillations at the natural frequency Examples singing wine glasses (stick-slip friction) Tacoma Narrows bridge (wind eddies) CP 466
  18. 18. 18 CP 466
  19. 19. 19 What is a good strategy for answering examination questions ???
  20. 20. 201 Read and answer the question 2 Type of problem Identify the physics – what model can be used? Use exam formula sheet 3 Answer in point form Break the question into small parts, do step by step showing all working and calculations (if can’t get a number in early part of a question, use a “dummy” number. Explicit physics principles (justification, explanation) Annotated diagrams (collect and information & data – implicit + explicit Equations Identify ⇔ Setup ⇔ Execute ⇔ Evaluate
  21. 21. 21 Problem 4.1 Why do some tall building collapse during an earthquake ?
  22. 22. 22I S E E Vibration motion can be resolved into vertical and horizontal motions
  23. 23. 23 Vertical motion m k 1 2 o k f mπ = natural frequency of vibration driving frequency fd 0 0.5 1 1.5 2 2.5 3 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 amplitudeA(m) ωd /ωo b = 2 b = 8 b = 10 Resonance fd → fo large amplitude oscillations – building collapses
  24. 24. 24 Driver frequency fd nodes antinodes 2nd floor disappeared – driving frequency matches natural frequency (3rd harmonic) Horizontal Motion Resonance Standing Waves setup in building 2nd floor
  25. 25. 25 Problem 4.2 Consider a tractor driving across a field that has undulations at regular intervals. The distance between the bumps is about 4.2 m. Because of safety reasons, the tractor does not have a suspension system but the driver’s seat is attached to a spring to absorb some of the shock as the tractor moves over rough ground. Assume the spring constant to be 2.0×104 N.m-1 and the mass of the seat to be 50 kg and the mass of the driver, 70 kg. The tractor is driven at 30 km.h-1 over the undulations. Will an accident occur?
  26. 26. 26 ∆x = 4.2 m k = 2x104 N.m-1 v = 30 km.h-1 m = (50 + 70) kg = 120 kg Solution 4.2 I S E E
  27. 27. 27 Tractor speed v = ∆x / ∆t = 30 km.h-1 = (30)(1000) / (3600) m.s-1 = 8.3 m.s-1 The time interval between hitting the bumps (∆x = 4.2 m) ∆t = ∆x / v = (4.2 / 8.3) s = 0.51 s Therefore, the frequency at which the tractor hits the bumps and energy is supplied to the oscillating system of spring-seat-person f = 1 / ∆t = 1 / 0.51 = 2.0 Hz. The natural frequency of vibration of the spring-seat-person is 4 2. 1 1 2 10 2 2 120 1 Hz k f mπ π × = = = This is an example of forced harmonic motion. Since the driving frequency (due to hitting the bumps) is very close to the natural frequency of the spring-seat- person the result will be large amplitude oscillations of the person and which may lead to an unfortunate accident. If the speed of the tractor is reduced, the driving frequency will not match the natural frequency and the amplitude of the vibration will be much reduced.

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