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Chapter 14 - Simple Harmonic Motion

Chapter 14 - Simple Harmonic Motion. A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University. © 2007. Photo by Mark Tippens.

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Chapter 14 - Simple Harmonic Motion

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  1. Chapter 14 - Simple Harmonic Motion A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University © 2007

  2. Photo by Mark Tippens A TRAMPOLINE exerts a restoring force on the jumper that is directly proportional to the average force required to displace the mat. Such restoring forces provide the driving forces necessary for objects that oscillate with simple harmonic motion.

  3. Objectives: After finishing this unit, you should be able to: • Write and apply Hooke’s Law for objects moving with simple harmonic motion. • Write and apply formulas for finding the frequency f,period T,velocity v, or accelerationa in terms of displacementxor time t. • Describe the motion of pendulums and calculate the length required to produce a given frequency.

  4. AmplitudeA Periodic Motion Simple periodic motion is that motion in which a body moves back and forth over a fixed path, returning to each position and velocity after a definite interval of time. Period, T, is the time for one complete oscillation. (seconds,s) Frequency, f, is the number of complete oscillations per second. Hertz (s-1)

  5. x F Example 1:The suspended mass makes 30 complete oscillations in 15 s. What is the period and frequency of the motion? Period: T = 0.500 s Frequency: f = 2.00 Hz

  6. x F Simple Harmonic Motion, SHM Simple harmonic motionis periodic motion in the absence of friction and produced by a restoring force that is directly proportional to the displacement and oppositely directed. A restoring force, F, actsin the direction opposite the displacement of the oscillating body. F = -kx

  7. x F m DF Dx k = Hooke’s Law When a spring is stretched, there is a restoring force that is proportional to the displacement. F = -kx The spring constant k is a property of the spring given by:

  8. x F m F (x) = kx F x1 x2 Work Done in Stretching a Spring Work done ONthe spring is positive; work BY spring is negative. From Hooke’s law the force F is: To stretch spring from x1 to x2 , work is: (Review module on work)

  9. 20 cm F m DF Dx 39.2 N 0.2 m k = = Example 2:A 4-kg mass suspended from a spring produces a displacement of 20 cm. What is the spring constant? The stretching force is the weight (W = mg) of the 4-kg mass: F = (4 kg)(9.8 m/s2) = 39.2 N Now, from Hooke’s law, the force constant k of the spring is: k = 196 N/m

  10. 8 cm F 0 m Example 2(cont.:The mass m is now stretched a distance of 8 cm and held. What is the potential energy? (k = 196 N/m) The potential energy is equal to the work done in stretching the spring: U = 0.627 J

  11. x m x = -A x = 0 x = +A Displacement in SHM • Displacement is positive when the position is to the right of the equilibrium position (x = 0) and negative when located to the left. • The maximum displacement is called the amplitude A.

  12. v (-) v (+) m x = -A x = 0 x = +A Velocity in SHM • Velocity is positive when moving to the right and negative when moving to the left. • It is zero at the end points and a maximum at the midpoint in either direction (+ or -).

  13. +a -a -x +x m x = -A x = 0 x = +A Acceleration in SHM • Acceleration is in the direction of the restoring force. (a is positive when x is negative, and negative when x is positive.) • Acceleration is a maximum at the end points and it is zero at the center of oscillation.

  14. a v x m x = -A x = 0 x = +A Acceleration vs. Displacement Given the spring constant, the displacement, and the mass, the acceleration can be found from: or Note: Acceleration is always opposite to displacement.

  15. a +x m Example 3:A 2-kg mass hangs at the end of a spring whose constant is k = 400 N/m. The mass is displaced a distance of 12 cm and released. What is the acceleration at the instant the displacement is x = +7 cm? a = -14.0 m/s2 Note: When the displacement is +7 cm (downward), the acceleration is -14.0 m/s2 (upward) independent of motion direction.

  16. +x m Example 4:What is the maximum acceleration for the 2-kg mass in the previous problem? (A = 12 cm, k = 400 N/m) The maximum acceleration occurs when the restoring force is a maximum; i.e., when the stretch or compression of the spring is largest. F = ma = -kx xmax =  A Maximum Acceleration: amax = ± 24.0 m/s2

  17. a v x m x = -A x = 0 x = +A Conservation of Energy The total mechanical energy(U + K) of a vibrating system is constant; i.e., it is the same at any point in the oscillating path. For any two points A and B, we may write: ½mvA2 + ½kxA2 = ½mvB2 + ½kxB2

  18. m a v x x = -A x = 0 x = +A Energy of a Vibrating System: A B • At points A and B, the velocity is zero and the acceleration is a maximum. The total energy is: U + K = ½kA2x =  A and v = 0. • At any other point: U + K = ½mv2 + ½kx2

  19. m a v x x = -A x = 0 x = +A vmaxwhen x = 0: Velocity as Function of Position.

  20. +x m Example 5:A 2-kg mass hangs at the end of a spring whose constant is k = 800 N/m. The mass is displaced a distance of 10 cm and released. What is the velocity at the instant the displacement is x = +6 cm? ½mv2 + ½kx 2 = ½kA2 v = ±1.60 m/s

  21. 0 +x m Example 5 (Cont.):What is the maximum velocity for the previous problem? (A = 10 cm, k = 800 N/m, m = 2 kg.) The velocity is maximum when x = 0: ½mv2 + ½kx 2 = ½kA2 v = ± 2.00 m/s

  22. w = 2f The Reference Circle The reference circle compares the circular motion of an object with its horizontal projection. x = Horizontal displacement. A = Amplitude (xmax). q = Reference angle.

  23. Velocity in SHM The velocity (v) of an oscillating body at any instant is the horizontal component of its tangential velocity (vT). vT = wR = wA; w = 2f v = -vTsin ;  = wt v = -wA sin w t v = -2f A sin 2ft

  24. Acceleration Reference Circle The acceleration(a) of an oscillating body at any instant is the horizontal component of its centripetal acceleration (ac). a = -accos q = -ac cos(wt) R = A a = -w2A cos(wt)

  25. The Period and Frequency as a Function of a and x. For any body undergoing simple harmonic motion: Since a = -4p2f2x and T = 1/f The frequency and the period can be found if the displacement and acceleration are known. Note that the signs of aand x will always be opposite.

  26. Period and Frequency as a Function of Mass and Spring Constant. For a vibrating body with an elastic restoring force: Recall that F = ma = -kx: The frequencyfand the periodTcan be found if the spring constant k and mass m of the vibrating body are known. Use consistent SI units.

  27. m a v x x = 0 x = -0.2 m x = +0.2 m Example 6:The frictionless system shown below has a 2-kg mass attached to a spring (k = 400 N/m). The mass is displaced a distance of 20 cm to the right and released.What is the frequency of the motion? f = 2.25 Hz

  28. m a v x x = 0 x = -0.2 m x = +0.2 m Example 6 (Cont.):Suppose the 2-kg mass of the previous problem is displaced 20 cm and released (k = 400 N/m). What is the maximum acceleration? (f = 2.25 Hz) Acceleration is a maximum when x =  A a =  40 m/s2

  29. m a v x x = 0 x = -0.2 m x = +0.2 m Example 6:The 2-kg mass of the previous example is displaced initially at x = 20 cm and released. What is the velocity 2.69 s after release? (Recall that f = 2.25 Hz.) v = -2f A sin 2ft (Note: q in rads) The minus sign means it is moving to the left. v = -0.916 m/s

  30. -0.12 m m a v x x = 0 x = -0.2 m x = +0.2 m Example 7:At what time will the 2-kg mass be located 12 cm to the left of x = 0? (A = 20 cm, f = 2.25 Hz) t = 0.157 s

  31. L mg The Simple Pendulum The period of a simple pendulum is given by: For small angles q.

  32. L L = 0.993 m Example 8. What must be the length of a simple pendulum for a clock which has a period of two seconds (tick-tock)?

  33. The Torsion Pendulum The period T of a torsion pendulum is given by: Where k’ is a torsion constant that depends on the material from which the rod is made; I is the rotational inertia of the vibrating system.

  34. Example 9:A 160 g solid disk is attached to the end of a wire, then twisted at 0.8 rad and released. The torsion constant k’ is 0.025 N m/rad. Find the period. (Neglect the torsion in the wire) For Disk: I = ½mR2 I = ½(0.16 kg)(0.12 m)2 =0.00115 kg m2 T = 1.35 s Note: Period is independent of angular displacement.

  35. x F m Summary Simple harmonic motion (SHM) is that motion in which a body moves back and forth over a fixed path, returning to each position and velocity after a definite interval of time. The frequency (rev/s) is the reciprocal of the period (time for one revolution).

  36. x F m Summary (Cont.) Hooke’s Law: In a spring, there is a restoringforce that is proportional to the displacement. The spring constant k is defined by:

  37. m a v x x = -A x = 0 x = +A Summary (SHM) Conservation of Energy: ½mvA2 + ½kxA2 = ½mvB2 + ½kxB2

  38. Summary (SHM)

  39. m a v x x = -A x = 0 x = +A Summary: Period and Frequency for Vibrating Spring.

  40. L Summary: Simple Pendulum and Torsion Pendulum

  41. CONCLUSION: Chapter 14Simple Harmonic Motion

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