2. Physics 151: Lecture 31, Pg 1 Physics 151: Lecture 31 Today’s Agenda l Today’s Topics çSimple Harmonic Motion – masses on springs (Ch 15.1-2) çEnergy of the SHO – Ch. 15.3 çPendula.

3. Physics 151: Lecture 31, Pg 2 Simple Harmonic Motion (SHM) l 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). l This oscillation is called Simple Harmonic Motion, and is actually very easy to understand... k m k m k m See text: 15.1-2

4. Physics 151: Lecture 31, Pg 3 SHM Dynamics Fa l At any given instant we know that F = ma must be true. l But in this case F = -kx and ma = l So: -kx = ma = k x m F F = -kxa a differential equation for x(t) ! See text: 15.1-2

5. Physics 151: Lecture 31, Pg 4 SHM Dynamics... Try the solution x = Acos(  t) this works, so it must be a solution ! define See text: 15.1-2

6. Physics 151: Lecture 31, Pg 5 SHM Dynamics... y = R cos  = R cos (  t ) l But wait a minute...what does angular frequency  have to do with moving back & forth in a straight line ?? x y 1 0  1 1 22 33 4 4 55 66 See text: 15.1-2

7. Physics 151: Lecture 31, Pg 6 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 See text: 15.1-2

8. Physics 151: Lecture 31, Pg 7 SHM Solution... Drawing of Acos(  t ) l A = amplitude of oscillation    T = 2  /  A A See text: 15.1-2

9. Physics 151: Lecture 31, Pg 8 SHM Solution... Drawing of Acos(  t +  )     See text: 15.1-2

10. Physics 151: Lecture 31, Pg 9 SHM Solution... Drawing of Acos(  t -  /2) A     = Asin(  t) ! See text: 15.1-2

11. Physics 151: Lecture 31, Pg 10 Velocity and Acceleration k x m 0 Position: x(t) = Acos(  t +  ) Velocity: v(t) = -  Asin(  t +  ) Acceleration: a(t) = -  2 Acos(  t +  ) See text: 15.2 by taking derivatives, since: x MAX = A v MAX =  A a MAX =  2 A

12. Physics 151: Lecture 31, Pg 11 Lecture 31, Act 1 Simple Harmonic Motion l 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 ? t y(t) (a) (b) (c)

13. Physics 151: Lecture 31, Pg 12 Lecture 31, Act 1 Solution l The slope of y(t) tells us the sign of the velocity since t y(t) (a) (b) (c) y(t) and a(t) have the opposite sign since a(t) = -  2 y(t) a < 0 v < 0 a > 0 v > 0 a 0 The answer is (c).

14. Physics 151: Lecture 31, Pg 13 Example l 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  ? çWhat is the spring constant k ? k x m See text: 15.2  = Also: k = m  2 So k = (2 kg) x (20 s -1 ) 2 = 800 kg/s 2 = 800 N/m v MAX =  A

15. Physics 151: Lecture 31, Pg 14 Initial Conditions k x m 0 Use “initial conditions” to determine phase  ! Suppose we are told x(0) = 0, and x is initially increasing (i.e. v(0) = positive): x(t) = Acos(  t +  ) v(t) = -  Asin(  t +  ) a(t) = -  2 Acos(  t +  )  sin cos  See text: 15.2  = -  /2 So

16. Physics 151: Lecture 31, Pg 15 Lecture 31, Act 2 Initial Conditions l 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): k m y 0 d (c) v(t) = - v max sin(  t) a(t) = -a max cos(  t) (a) v(t) = v max sin(  t) a(t) = a max cos(  t) (b) v(t) = v max cos(  t) a(t) = -a max cos(  t) (both v max and a max are positive numbers) t = 0

17. Physics 151: Lecture 31, Pg 16 Energy of the Spring-Mass System We know enough to discuss the mechanical energy of the oscillating mass on a spring. Kinetic energy is always K = 1/2 mv 2 K = 1/2 m (-  Asin(  t +  )) 2 We also know what the potential energy of a spring is, U = 1/2 k x 2 U = 1/2 k (Acos(  t +  )) 2 x(t) = Acos(  t +  ) v(t) = -  Asin(  t +  ) a(t) = -  2 Acos(  t +  ) See text: 15.2 Remember,

18. Physics 151: Lecture 31, Pg 17 Energy of the Spring-Mass System Add to get E = K + U 1/2 m (  A) 2 sin 2 (  t +  ) + 1/2 k (Acos(  t +  )) 2 Remember that    See text: 15.3 U~cos 2 K~sin 2 E = 1/2 kA 2 so, E = 1/2 kA 2 sin 2 (  t +  ) + 1/2 kA 2 cos 2 (  t +  ) = 1/2 kA 2 ( sin 2 (  t +  ) + cos 2 (  t +  )) = 1/2 kA 2

19. Physics 151: Lecture 31, Pg 18 SHM So Far The most general solution is x = Acos(  t +  ) where A = amplitude  = frequency  = phase constant l For a mass on a spring çThe frequency does not depend on the amplitude !!! çWe will see that this is true of all simple harmonic motion ! l The oscillation occurs around the equilibrium point where the force is zero! See text: 15.1 to 15.3

20. Physics 151: Lecture 31, Pg 19 The Simple Pendulum l 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.  L m mg z See text: 15.4

21. Physics 151: Lecture 31, Pg 20 Aside: sin  and cos  for small  l A Taylor expansion of sin  and cos  about  = 0 gives: and So for  <<1, and

22. Physics 151: Lecture 31, Pg 21 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  =  mL 2  L d m mg z where Differential equation for simple harmonic motion !  =  0 cos(  t +  ) See text: 15-5

23. Physics 151: Lecture 31, Pg 22 Lecture 31, Act 3 Simple Harmonic Motion l You are sitting on a swing. A friend gives you a small push and you start swinging back & forth with period T 1. l Suppose you were standing on the swing rather than sitting. When given a small push you start swinging back & forth with period T 2. çWhich of the following is true: (a) T 1 = T 2 (b) T 1 > T 2 (c) T 1 < T 2

24. Physics 151: Lecture 31, Pg 23 Lecture 31, Act 3 Solution L1L1 L2L2 Standing up raises the CM of the swing, making it shorter ! T1T1 T2T2 Since L 1 > L 2 we see that T 1 > T 2. (b)

25. Physics 151: Lecture 31, Pg 24 Lecture 31, Act 4 Simple Harmonic Motion l 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.

26. Physics 151: Lecture 31, Pg 25 The Rod Pendulum l 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.  L mg z x CM See text: 15.4

27. Physics 151: Lecture 31, Pg 26 The Rod Pendulum... The torque about the rotation (z) axis is  = -mgd = -mg{L/2}sin  -mg{L/2}  for small  l In this case  L d mg z L/2 x CM where See text: 15.4 d I So  = I  becomes

28. Physics 151: Lecture 31, Pg 27 Lecture 31, Act 4 Period (a)(b)(c) l What length do we make the simple pendulum so that it has the same period as the rod pendulum? LRLR LSLS

29. Physics 151: Lecture 31, Pg 28 b)  S =  R if See text: 15-5 Lecture 31, Act 4 Solution LSLS LRLR

30. Physics 151: Lecture 31, Pg 29 Recap of today’s lecture l Simple Harmonic Motion, çExample, block on a spring çEnergy of SHM çPendula are just like block/spring