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Simple Harmonic Motion in Physics Lecture

Learn about simple harmonic motion in this physics lecture. Topics include masses on springs, energy of the SHO, and pendula. Understand the dynamics and solution of SHM with examples and equations.

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Simple Harmonic Motion in Physics Lecture

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  1. Physics 151: Lecture 31 Today’s Agenda • Today’s Topics • Simple Harmonic Motion – masses on springs (Ch 15.1-2) • Energy of the SHO – Ch. 15.3 • Pendula.

  2. k m k m k m See text: 15.1-2 Simple Harmonic Motion (SHM) • We know that if we stretch a spring with a mass on the end and let it go the mass will oscillate back and forth (if there is no friction). • This oscillation is called Simple Harmonic Motion,and is actually very easy to understand...

  3. F = -kx a k m x See text: 15.1-2 SHM Dynamics • At any given instant we know thatF = mamust be true. • But in this case F = -kx and ma = • So: -kx = ma = a differential equation for x(t) !

  4. Try the solution x = Acos(t) See text: 15.1-2 SHM Dynamics... define this works, so it must be a solution !

  5. See text: 15.1-2 SHM Dynamics... • But wait a minute...what does angularfrequencyhave to do with moving back & forth in a straight line ?? y = Rcos =Rcos (t) y 1 1 1 2 2 3 3  0 x 4 6 -1 4 6 5 5

  6. See text: 15.1-2 SHM Solution • We just showed that (which came from F=ma) has the solution x = Acos(t) . • This is not a unique solution, though. x = Asin(t) is also a solution. • The most general solution is a linear combination of these two solutions! x = Bsin(t)+Ccos(t) • This is equivalent to: x = A cos(t+) where  is called a phase

  7. See text: 15.1-2 SHM Solution... • Drawing of Acos(t ) • A = amplitude of oscillation T = 2/ A     A

  8. See text: 15.1-2 SHM Solution... • Drawing of Acos(t + )     

  9. See text: 15.1-2 SHM Solution... • Drawing of Acos(t - /2)  A     = Asin(t) !

  10. by taking derivatives, since: xMAX = A vMAX = A aMAX = 2A k m x 0 See text: 15.2 Velocity and Acceleration Position: x(t) = Acos(t + ) Velocity: v(t) = -Asin(t + ) Acceleration: a(t) = -2Acos(t + )

  11. Lecture 31, Act 1Simple Harmonic Motion • A mass oscillates up & down on a spring. It’s position as a function of time is shown below. At which of the points shown does the mass have positive velocity and negative acceleration ? y(t) (a) (c) t (b)

  12. Lecture 31, Act 1Solution • The slope of y(t) tells us the sign of the velocity since • y(t) and a(t) have the opposite sign since a(t) = -w2 y(t) a < 0v > 0 a < 0v < 0 y(t) (a) (c) t (b) a > 0v > 0 The answer is (c).

  13. vMAX =A  = Also: k = m2 k m x See text: 15.2 Example • A mass m = 2kg on a spring oscillates with amplitude A = 10cm. At t=0 its speed is maximum, and is v = +2 m/s. • What is the angular frequency of oscillation  ? • What is the spring constant k ? Sok = (2 kg) x (20 s -1) 2 = 800 kg/s2 = 800 N/m

  14. x(t) = Acos(t + ) v(t) = -Asin(t + ) a(t) = -2Acos(t + ) So  = -/2    k sin cos m x 0 See text: 15.2 Initial Conditions Use “initial conditions” to determine phase ! Suppose we are told x(0) = 0 , and x is initially increasing (i.e. v(0) = positive):

  15. m Lecture 31, Act 2Initial Conditions • A mass hanging from a vertical spring is lifted a distance d above equilibrium and released at t = 0. Which of the following describe its velocity and acceleration as a function of time (upwards is positive y direction): (a) v(t) = vmax sin(wt) a(t) = amax cos(wt) k y (b) v(t) = vmax cos(wt) a(t) = -amax cos(wt) d t = 0 (c) v(t) = - vmax sin(wt) a(t) = -amax cos(wt) 0 (both vmax and amax are positive numbers)

  16. x(t) = Acos(t + ) v(t) = -Asin(t + ) a(t) = -2Acos(t + ) See text: 15.2 Energy of the Spring-Mass System We know enough to discuss the mechanical energy of the oscillating mass on a spring. Remember, Kinetic energy is always K = 1/2 mv2 K = 1/2 m (-Asin(t + ))2 We also know what the potential energy of a spring is, U = 1/2 k x2 U = 1/2 k (Acos(t + ))2

  17. E = 1/2 kA2 U~cos2 K~sin2    See text: 15.3 Energy of the Spring-Mass System Add to get E = K + U 1/2 m (A)2sin2(t + ) + 1/2 k (Acos(t + ))2 Remember that so, E = 1/2 kA2 sin2(t + ) + 1/2 kA2 cos2(t + ) = 1/2 kA2 ( sin2(t + ) + cos2(t + )) = 1/2 kA2

  18. See text: 15.1 to 15.3 SHM So Far • The most general solution is x = Acos(t + ) where A = amplitude  = frequency  = phase constant • For a mass on a spring • The frequency does not depend on the amplitude !!! • We will see that this is true of all simple harmonic motion ! • The oscillation occurs around the equilibrium point where the force is zero!

  19. z  L m mg See text: 15.4 The Simple Pendulum • A pendulum is made by suspending a mass m at the end of a string of length L. Find the frequency of oscillation for small displacements.

  20. Aside: sin and cos  for small  • A Taylor expansion of sin  and cos about  = 0gives: and So for <<1, and

  21. z  L where m Differential equation for simple harmonic motion ! d  = 0 cos(t + ) mg See text: 15-5 The Simple Pendulum... • Recall that the torque due to gravity about the rotation (z) axis is = -mgd. d = Lsin   L for small so = -mg L • But =II=mL2

  22. Lecture 31, Act 3Simple Harmonic Motion • You are sitting on a swing. A friend gives you a small push and you start swinging back & forth with period T1. • Suppose you were standing on the swing rather than sitting. When given a small push you start swinging back & forth with period T2. • Which of the following is true: (a)T1 = T2(b)T1 > T2(c) T1 < T2

  23. (b) Since L1 > L2 we see that T1 > T2 . Lecture 31, Act 3Solution Standing up raises the CM of the swing, making it shorter ! L2 L1 T1 T2

  24. Lecture 31, Act 4Simple Harmonic Motion • Two clocks with basic timekeeping mechanism consist of 1) a mass on a string and 2) a simple pendulum. Both have a period of 1s on Earth. When taken to the moon which one of the statements below is correct ? a) the periods of both is unchanged. b) one of them has a period shorter than 1 s. c) the pendulum has a period longer than 1 s. d) the mass-spring system has a period longer than 1s. e) both c) and d) are true.

  25. See text: 15.4 The Rod Pendulum • A pendulum is made by suspending a thin rod of length L and mass M at one end. Find the frequency of oscillation for small displacements. z  x CM L mg

  26. d I • So =Ibecomes where See text: 15.4 The Rod Pendulum... • The torque about the rotation (z) axis is= -mgd= -mg{L/2}sinq -mg{L/2}q for small q • In this case z L/2  x CM L d mg

  27. LS LR Lecture 31, Act 4Period • What length do we make the simple pendulum so that it has the same period as the rod pendulum? (a)(b) (c)

  28. LS LR b) S = Rif See text: 15-5 Lecture 31, Act 4Solution

  29. Recap of today’s lecture • Simple Harmonic Motion, • Example, block on a spring • Energy of SHM • Pendula are just like block/spring

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