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Exploring the Similarities and Differences Between Graphs in Oscillatory Motion

Delve into the world of oscillatory motion as we examine the characteristics of graphs related to mass on spring displacement, periodic motion, and simple harmonic motion. Understand the nuances of displacement, velocity, and acceleration through insightful explanations and examples.

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Exploring the Similarities and Differences Between Graphs in Oscillatory Motion

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  1. What are the similarities and differences between graphs? Mass on spring Displacement Both show oscillatory or cyclic motion

  2. Oscillatory/Periodic MotionRepetitive Motion with a period and frequency. • Motion can be driven internally - mass on spring. • Externally – pendulum or tides.

  3. Guitar string EEG Beta Waves

  4. SHM – special case • What do you notice?

  5. Vocabulary • Period T: time required for one cycle of periodic motion (sec). • Frequency: number of oscillations per unit time. unit is Hertz:

  6. What are the period and frequency of the ECG • T = 1 sec. • f = 1 hz,

  7. Angular Frequency - w (rad/sec) • w = 2pf. • remember angular speed from circular motion? • w = Dq/t • w = 2 p rad/s • Equilibrium (0): the spot the mass would come to rest when not disturbed – Fnet = 0. • Displacement: (s or x): distance from equilibrium. • Amplitude (xo) – max displacement from eq.

  8. Representation of Oscillatory Motion • Observe the motion of a bobbing mass. Where is the: • Displacement positive? • Displacement negative? • Displacement zero? • velocity positive? • Where is the velocity negative? • Where is the velocity zero?

  9. Simple Harmonic Motion – SHM • Isochronous period. • Restoring Force directly proportional to displacement. Double displacement, double force etc. • Further from equilibrium, more force directed toward it. • Force and displacement in opposite directions.

  10. Pendulum is not SHM • Fnet not directly opposite s • for small displacement angles it approximates SHM.

  11. Free Body Diagram Mass on Spring • Remember Hooke’s Law? • F = -kx. • F is the restoring force of the spring. • Complete sheet.

  12. Hwk Intro

  13. Simple Harmonic Motion2 Conditions. • 1. Acceleration/Fnet proportional to displacement. • 2. Acceleration/Fnet directed toward equilibrium. • Defining Equation for SHM • a = -w2x

  14. Graphs Of SHM

  15. Acceleration - displacement • Since F = - kx. • ma = - kx. • a = -k x • m • a a – x • Graph a (Y axis) vs. displacement on the X

  16. Negative gradient = when displacement is positive, acceleration and restoring force are negative • ∴ a ∝ − x • experimental evidence shows k = ω2, where ω is the angular velocity, which is be ω = 2Πf : • a =  − ω2x a =  − kx

  17. v = 0 v = vMAX v = 0 Sketching and interpreting graphs of SHM x -2.0 0.0 2.0 EXAMPLE: The displacement x vs. time t for a system undergoing SHM is shown here. Sketch in red the velocity vs. time graph. SOLUTION: At the extremes, v = 0. At x = 0, v = vMAX. The slope determines sign of vMAX. x-black v-red (+) (-) (+) (-) (+) t

  18. v = 0 v = vMAX v = 0 x Sketching and interpreting graphs of simple harmonic motion -2.0 0.0 2.0 EXAMPLE: The displacement x vs. time t for a system undergoing SHM is shown here. Sketch in blue the acceleration vs. time graph. SOLUTION: Since a -x, a is just a reflection of x. Note: x is a sine, v is a cosine, and a is a – sine wave. x-black v-red (different scale) t a-blue (different scale)

  19. Plot of displacement, velocity, acceleration, on same graph.

  20. Phase and Phase DifferenceTwo waves in Phase

  21. Phase and Phase DifferencePhase shift = p

  22. 4. What is the phase shift?

  23. Phase Difference • The graphs all have the same period, but • velocity leads displacement by ¼ of a period 90o. • Acceleration leads velocity by ¼ of a period 90o. • Displacement is 90o out of phase with velocity and 180o out of phase with acceleration. • When the phase difference is 0o or 360o, the systems are “in phase”.

  24. x -2.0 0.0 2.0 EXAMPLE: The kinetic energy vs. displacement for a system undergoing SHM is shown in the graph. The system consists of a 0.125-kg mass on a spring. (a) Determine the maximum velocity of the mass. SOLUTION: When the kinetic energy is maximum, the velocity is also maximum. Thus 4.0 = (1/2)mvMAX2 so that 4.0 = (1/2)(.125)vMAX2 vMAX = 8.0 ms-1.

  25. x -2.0 0.0 2.0 ET EXAMPLE: The kinetic energy vs. displacement for a system undergoing SHM is shown in the graph. The system consists of a 0.125-kg mass on a spring. (b) Sketch EP and determine the total energy of the system. SOLUTION: Since EK + EP = ET = CONST, and since EP = 0 when EK = EK,MAX, it must be that ET = EK,MAX = 4.0 J. Thus the EP graph will be the “inverted” EK graph. EK EP

  26. x -2.0 0.0 2.0 EXAMPLE: The kinetic energy vs. displacement for a system undergoing SHM is shown in the graph. The system consists of a 0.125-kg mass on a spring. (c) Determine the spring constant k of the spring. SOLUTION: Recall EP = (1/2)kx2. Note that EK = 0 at x = xMAX = 2.0 cm. Thus EK + EP = ET = CONSTET = 0 + (1/2)kxMAX2 so that 4.0 = (1/2)k 0.0202  k = 20000 Nm-1.

  27. x -2.0 0.0 2.0 EXAMPLE: The kinetic energy vs. displacement for a system undergoing SHM is shown in the graph. The system consists of a 0.125-kg mass on a spring. (d) Determine the acceleration of the mass at x = 1.0 cm. SOLUTION: From Hooke’s law, F = -kx we get F = -20000(0.01) = -200 N. From F = ma we get -200 = 0.125a  a = -1600 ms-2.

  28. x -2.0 0.0 2.0 EXAMPLE: A 4.0-kg mass is placed on a spring’s end and displaced 2.0 m to the right. The spring force F vs. its displacement x from equilibrium is shown in the graph. (a) How do you know that the mass is undergoing SHM? SOLUTION: In SHM, a -x. Since F = ma, then F  -x also. The graph shows that F  -x. Thus we have SHM.

  29. x -2.0 0.0 2.0 F = -5.0 N x = 1.0 m EXAMPLE: A 4.0-kg mass is placed on a spring’s end and displaced 2.0 m to the right. The spring force F vs. its displacement x from equilibrium is shown in the graph. (b) Find the spring constant of the spring. SOLUTION: Use Hooke’s law: F = -kx. Pick any F and any x. Use k = -F / x. Thus k = -(-5.0 N) / 1.0 m = 5.0 Nm-1.

  30. x -2.0 0.0 2.0 EXAMPLE: A 4.0-kg mass is placed on a spring’s end and displaced 2.0 m to the right. The spring force F vs. its displacement x from equilibrium is shown in the graph. (c) Find the total energy of the system. SOLUTION: Use ET = (1/2)kxMAX2. Then ET = (1/2)kxMAX2= (1/2)  5.0  2.02 = 10. J.

  31. x -2.0 0.0 2.0 EXAMPLE: A 4.0-kg mass is placed on a spring’s end and displaced 2.0 m to the right. The spring force F vs. its displacement x from equilibrium is shown in the graph. (d) Find the maximum speed of the mass. SOLUTION: Use ET = (1/2)mvMAX2. 10. = (1/2)  4.0  vMAX2 vMAX = 2.2 ms-1.

  32. x -2.0 0.0 2.0 EXAMPLE: A 4.0-kg mass is placed on a spring’s end and displaced 2.0 m to the right. The spring force F vs. its displacement x from equilibrium is shown in the graph. (e) Find the speed of the mass when its displacement is 1.0 m. SOLUTION: Use ET = (1/2)mv 2 + (1/2)kx 2. Then 10. = (1/2)(4)v 2+ (1/2)(5)12 v = 1.9 ms-1.

  33. Graphical Treatment Equations of SHM

  34. Displacement, x, against time x = xo cos wt start point at max ampl. ** Set Calculator in Radians.

  35. Displacement against time x = xosinwt • Speed = d/t = 2pR/T but • 2p /T = w so • v = wr but r the displacement x so. • v = wx.

  36. Velocity against time v = vocoswt Midpoint = max velocity. Starting where?

  37. Equations of Graphs • x = xocoswt x = xo sin wt • v = -vo sin wt v = vocoswt • a = -aocoswt -aosin wt • Released from Released equilibrium. top.

  38. Ex 2. A mass on a spring is oscillating with f = 0.2 Hz and xo = 3 cm. What is the displacement of the mass 10.66 s after its release from the top? • x = xo cos wt xo = 3 cm • w = 2pf. = 0.4 p Hz =1.26 rad/s. • t = 10.66 s • x = 0.03 cos (1.26 x 10.66) = 0.019 m • You must use radians on calculator.

  39. SHM and Circular Motion • Use the relationship to derive equations.

  40. If an object moving with constant speed in a circular path is observed from a distant point (in the plane of the motion), it will appear to be oscillating with SHM. The shadow of a pendulum bob moves with s.h.m. when the pendulum itself is either oscillating (through a small angle) or moving in a circle with constant speed, as shown in the diagram.

  41. For any s.h.m. we can find a corresponding circular motion. When a circular motion "corresponds to" a given s.h.m., i) the radius of the circle is equal to the amplitude of the s.h.m.  ii) the time period of the circular motion is equal to the time period of the s.h.m.

  42. Derive Relationship between accl & w for SHM. From circular motion ac = v2/r and vc = 2pr/TOscillating systems have acceleration too. • But w = 2p/T • vc = 2pr/T • vc = wr • But ac = v2/r • So ac = (wr)2/r • but r is related to displacement x.

  43. For any displacement: a = -w²x ao = -w²xo The negative sign shows Fnet & accl direction opposite displacement. Derivation of accl in Hamper pg 76. Or use 2nd derivative of displacement.

  44. Ex 3. A pendulum swings with f = 0.5 Hz. What is the size & direction of the acceleration when the bob has displacement of 2 cm right? • a = -w²x • w = 2pf = p • a = -(p)2 (0.02 m) = -0.197 m/s2. left.

  45. Ex 4: A mass is bobbing on a spring with a period of 0.20 seconds. What is its angular acceleration at a point where its displacement is 1.5 cm? • w = 2p/T • .w = 31 rad/s • a = -w²x • a = (31rad/s)(1.5 cm) = 1480 cm/s2. • 15 m/s2.

  46. To find the velocity of an oscillating mass or pendulum at any displacement: When the mass is at equilibrium, x = 0, and velocity is maximum: vo = ± wxo. Derivation on H pg 77.

  47. Ex 5. A pendulum swings with f = 1 Hz and amplitude 3 cm. At what position will be its maximum velocity &what is the velocity? At max velocity vo = wxo. w = 2pf = 2p(1) = 2p rad/s vo = (2prad/s)(0.03) vo = 0.188 m/s vo = 0.2 m/s

  48. Hwk Hamper pg 75- 77 Show equations and work, hand in virtual solar system lab. Mechanical Universe w/questions • http://www.learner.org/resources/series42.html?pop=yes&pid=565

  49. SHM, Hooke’s Law & k.

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