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Oscillations

Oscillations. Unit 7. F s = -kx. (Hooke’s Law). Lesson 1 : Simple Harmonic Motion. F s is a restoring force because it always points toward the equilibrium position (x = 0). k. a x = -. x. m. Simple Harmonic Motion.

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Oscillations

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  1. Oscillations Unit 7

  2. Fs = -kx (Hooke’s Law) Lesson 1 : Simple Harmonic Motion Fs is a restoring force because it always points toward the equilibrium position (x = 0)

  3. k ax = - x m Simple Harmonic Motion An object moves with simple harmonic motion whenever its acceleration is proportional to its position and is oppositely directed to the displacement from equilibrium. Applying Newton’s Second Law : SF = max -kx = max

  4. _____ A/2 _____ A _____ 2A _____ 4A Example 1 A block on the end of a spring is pulled to position x = A and released. In one full cycle of its motion, through what total distance does it travel ?

  5. d2x k = - x dt2 m Notice that the acceleration of the particle in SHM is not constant. It varies with position x. d2x = -w2x dt2 dv d2x , Since a = = dt dt2 If we call k/m = w2,

  6. One possible solution is : x(t) = A cos(wt + f) d2x = -w2x dt2 (second-order differential equation) We need a function x(t) whose second derivative is the same as the original function with a negative sign and multiplied by w2.

  7. dx d = A cos(wt + f) = -wA sin(wt + f) dt dt d2x d = -wA sin(wt + f) = -w2A cos(wt + f) dt2 dt d2x = -w2x dt2 Proof

  8. k w = m x(t) = A cos(wt + f) Amplitude (A) : maximum value of the position of the particle in either the positive or negative direction. Angular Frequency (w) : number of oscillations per second. Since k/m = w2,

  9. If particle is at maximum position x = A at t = 0, the phase constant f = 0. f = 0 Phase Constant (f) : initial phase angle. This is determined by the position of the particle at t = 0.

  10. Pen traces out cosine curve x(t) = A cos(wt + f)

  11. 2p T = w Period (T) : the time interval required for the particle to go through one full cycle of its motion.

  12. 1 w f = f = T 2p Frequency (f) : the inverse of period. The number of oscillations that the particle undergoes per second. Since T = 2p/w, The Hertz (Hz) is the SI unit for frequency.

  13. 2p w = 2pf = T Since w = k/m and T = 2p/w , m T = 2p k depend only on mass and spring constant k 1 f = 2p m Since f = w/2p,

  14. dx d = A cos(wt + f) = -wA sin(wt + f) dt dt v = -wA sin(wt + f) d2x d = -wA sin(wt + f) = -w2A cos(wt + f) dt2 dt a = -w2A cos(wt + f) Velocity in Simple Harmonic Motion Acceleration in Simple Harmonic Motion

  15. (magnitude only) k vmax = k A amax = A (magnitude only) m m Maximum Speed in Simple Harmonic Motion Since sine and cosine oscillate between +/- 1, vmax = +/- wA Maximum Acc. in Simple Harmonic Motion amax = +/- w2A

  16. phase difference is p/2 rad or 90o phase difference is p rad or 180o position vs. time when x at max or min, v = 0 when x = 0, v is max velocity vs. time when x at max, a is max in opposite direction acceleration vs. time

  17. An object oscillates with simple harmonic motion along the x-axis. Its position varies with time according to the equation x = (4.00 m) cos(pt + p/4) where t is in seconds and the angles in the parentheses are in radians. Example 2 a) Determine the amplitude, frequency, and period of the motion.

  18. b) Calculate the velocity and acceleration of the object at any time t. c) Using the results of part b, determine the position, velocity, and acceleration of the object at t = 1.00 s.

  19. d) Determine the maximum speed and maximum acceleration of the object. e) Find the displacement of the object between t = 0 and t = 1.00 s.

  20. A 200 g block connected to a light spring for which the force constant is 5.00 N/m is free to oscillate on a horizontal, frictionless surface. The block is displaced 5.00 cm from equilibrium and released from rest, as shown below. Example 3 a) Find the period of its motion.

  21. b) Determine the maximum speed of the block. c) What is the maximum acceleration of the block ?

  22. d) Express the position, speed, and acceleration as functions of time. e) The block is released from the same initial position, xi = 5.00 cm, but with an initial velocity of vi = -0.100 m/s. Which parts of the solution change and what are the new answers for those that do change ?

  23. A 2 kg block is dropped from a height of 0.45 m above an uncompressed spring, as shown above. The spring has an elastic constant of 200 N/m and negligible mass. The block strikes the end of the spring and sticks to it. Example 4 : AP 1989 # 3

  24. a) Determine the speed of the block at the instant it hits the end of the spring. b) Determine the period of the simple harmonic motion that ensues.

  25. c) Determine the distance that the spring is compressed at the instant the speed of the block is maximum. d) Determine the maximum compression of the spring.

  26. e) Determine the amplitude of the simple harmonic motion.

  27. Example 5 : AP 2003 # 2 An ideal spring is hung from the ceiling and a pan of mass M is suspended from the end of the spring, stretching it a distance D as shown above. A piece of clay, also of mass M, is then dropped from a height H onto the pan and sticks to it. Express all algebraic answers in terms of the given quantities and fundamental constants.

  28. a) Determine the speed of the clay at the instant it hits the pan. b) Determine the speed of the pan just after the clay strikes it.

  29. c) Determine the period of the simple harmonic motion that ensues. d) Determine the distance the spring is stretched (from its initial unstretched length) at the moment the speed of the pan is a maximum. Justify your answer.

  30. e) Indicate below whether the period of the resulting simple harmonic motion of the pan is greater than, less than, or the same as it was in part c. ____ Greater than ____ Less than ____ The same as Justify your answer. The clay is now removed from the pan and the pan is returned to equilibrium at the end of the spring. A rubber ball, also of mass M, is dropped from the same height H onto the pan, and after the collision is caught in midair before hitting anything else.

  31. KE of Block KE = ½ mv2 Elastic PE Energy Stored in Spring U = ½ kx2 Lesson 2 : Energy in Simple Harmonic Motion Since v = -wA sin(wt + f), KE = ½ mw2A2 sin2(wt + f) Since x = A cos(wt + f), U = ½ kA2 cos2(wt + f)

  32. E = ½ kA2 Total Mechanical Energy of Simple Harmonic Oscillator E = KE + U E = ½ mw2A2 sin2(wt + f) + ½ kA2 cos2(wt + f) Since w2 = k/m, E = ½ kA2 [sin2(wt + f) + cos2(wt + f)] Since sin2q + cos2q = 1,

  33. E = ½ kA2 The total mechanical energy of a simple harmonic oscillator is a constant of the motion and is proportional to the square of the amplitude. U is small when KE is large, and vice versa. KE + U = constant

  34. k v = +/- (A2 – x2) m , velocity of simple harmonic oscillator k Since w = v = +/- w A2 – x2 m E = ½ kA2 Since E = KE + U, ½ kA2 = ½ mv2 + ½ kx2 Solving for v,

  35. Example 1 The amplitude of a system moving in simple harmonic motion is doubled. Determine the change in the a) total energy b) maximum speed

  36. c) maximum acceleration d) period

  37. Example 2 A 0.500 kg cart connected to a light spring for which the force constant is 20.0 N/m oscillates on a horizontal, frictionless air track. a) Calculate the total energy of the system and the maximum speed of the cart if the amplitude of the motion is 3.00 cm.

  38. b) What is the velocity of the cart when the position is 2.00 cm ? c) Compare the kinetic and potential energies of the system when the position is 2.00 cm ?

  39. Lesson 3 : Comparing Simple Harmonic Motion with Uniform Circular Motion As the turntable rotates with constant angular speed, the shadow of the ball moves back and forth in simple harmonic motion.

  40. t = 0 Reference Circle Simple harmonic motion along a straight line can be represented by the projection of uniform circular motion along a diameter of a reference circle.

  41. Since q = wt = f , x coordinate is x(t) = A cos(wt + f) Point Q moves with simple harmonic motion t > 0 Q is the projection of P

  42. What is the amplitude and phase constant (relative to an x axis to the right) of the simple harmonic motion of the ball’s shadow ? _____ 0.50 m and 0 _____ 1.00 m and 0 _____ 0.50 m and p _____ 1.00 m and p Example 1

  43. While riding behind a car traveling at 3.00 m/s, you notice that one of the car’s tires has a small hemispherical bump on its rim, as shown. Example 2 a) Explain why the bump, from your viewpoint behind the car, executes simple harmonic motion.

  44. b) If the radii of the car’s tires are 0.300 m, what is the bump’s period of oscillation ?

  45. A simple pendulum exhibits periodic motion. It consists of a particle-like bob of mass m suspended by a light string of length L that is fixed at the upper end. Lesson 4 : The Pendulum Forces acting on bob Tension in string Gravitational force mg

  46. Ft is a restoring force d2s Ft = -mg sinq = m dt2 d2q g = - sinq for small values of q dt2 L Ft = mg sinq always acts opposite to the displacement of the bob Since s = Lq and L is constant,

  47. same form d2x k = - x dt2 m For small amplitudes (q < about 10o), the motion of a pendulum is close to simple harmonic motion. d2q g = - sinq dt2 L

  48. q = qmax cos(wt + f) Instead of , g k angular frequency for a simple pendulum w = w = L m 2p L T = = 2p w g Instead of x = A cos(wt + f),

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