1. Physics 1501: Lecture 25, Pg 1 Physics 1501: Lecture 25 Today’s Agenda l Homework #9 (due Friday Nov. 4) l Midterm 2: Nov. 16 l Topics çReview of static equilibrium çOscillation çSimple Harmonic Motion – masses on springs çEnergy of the SHO

2. Physics 1501: Lecture 25, Pg 2 Approach to Statics: l In general, we can use the two equations to solve any statics problems. l When choosing axes about which to calculate torque, we can be clever and make the problem easy....

3. Physics 1501: Lecture 25, Pg 3 Lecture 25, Act 1 Statics l A box is placed on a ramp in the configurations shown below. Friction prevents it from sliding. The center of mass of the box is indicated by a white dot in each case. çIn which cases does the box tip over ? (a) all (b) 2 & 3 (c) 3 only 1 2 3

4. Physics 1501: Lecture 25, Pg 4 Lecture 25, Act 1 Solution l We have seen that the torque due to gravity acts as though all the mass of an object is concentrated at the center of mass. l Consider the bottom right corner of the box to be a pivot point. l If the box can rotate in such a way that the center of mass is lowered, it will !

5. Physics 1501: Lecture 25, Pg 5 Lecture 25, ACT 1 Solution l We have seen that the torque due to gravity acts as though all the mass of an object is concentrated at the center of mass. l Consider the bottom right corner of the box to be a pivot point. l If the box can rotate in such a way that the center of mass is lowered, it will !

6. Physics 1501: Lecture 25, Pg 6 Lecture 25, Act 1 Addendum l What are the torques ?? (where do the forces act ?) rGrG mgmg gg  switches sign at critical point rfrf f  f   always zero  goes to zero at critical point rNrN N NN

7. Physics 1501: Lecture 25, Pg 7 New topic (Ch. 13) 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

8. Physics 1501: Lecture 25, Pg 8 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) !

9. Physics 1501: Lecture 25, Pg 9 SHM Dynamics... Try the solution x = Acos(  t) this works, so it must be a solution ! define

10. Physics 1501: Lecture 25, Pg 10 SHM Solution l 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. l The most general solution is a linear combination of these two solutions! x = Bsin(  t)+ Ccos(  t) ok

11. Physics 1501: Lecture 25, Pg 11 Derivation: x = Acos(  t+  ) is equivalent to x = Bsin(  t)+ Ccos(  t) x = Acos(  t+  ) = Acos(  t) cos  - Asin(  t) sin  where C = Acos(  ) and B =  Asin(  ) It works! = Ccos(  t) + Bsin(  t) We want to use the most general solution: So we can use x = Acos(  t+  ) as the most general solution!

12. Physics 1501: Lecture 25, Pg 12 SHM Solution... Drawing of Acos(  t ) l A = amplitude of oscillation    T = 2  /  A A

13. Physics 1501: Lecture 25, Pg 13 SHM Solution... Drawing of Acos(  t +  )    

14. Physics 1501: Lecture 25, Pg 14 SHM Solution... Drawing of Acos(  t -  /2) A     = Asin(  t) !

15. Physics 1501: Lecture 25, Pg 15 What about Vertical Springs? l We already know that for a vertical spring if y is measured from the equilibrium position l The force of the spring is the negative derivative of this function: l So this will be just like the horizontal case: -ky = ma = j k m F= -ky y = 0 Which has solution y = Acos(  t +  ) where

16. Physics 1501: Lecture 25, Pg 16 Velocity and Acceleration k x m 0 Position: x(t) = Acos(  t +  ) Velocity: v(t) = -  Asin(  t +  ) Acceleration: a(t) = -  2 Acos(  t +  ) by taking derivatives, since: x MAX = A v MAX =  A a MAX =  2 A

17. Physics 1501: Lecture 25, Pg 17 Lecture 25, Act 2 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)

18. Physics 1501: Lecture 25, Pg 18 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  = Also: k = m  2 So k = (2 kg) x (20 s -1 ) 2 = 800 kg/s 2 = 800 N/m v MAX =  A

19. Physics 1501: Lecture 25, Pg 19 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  x(0) = 0 = Acos(  )  =  /2 or -  /2 v(0) > 0 = -  Asin(  )  < 0  = -  /2 So

20. Physics 1501: Lecture 25, Pg 20 Initial Conditions... k x m 0 x(t) = Acos(  t -  /2 ) v(t) = -  Asin(  t -  /2 ) a(t) = -  2 Acos(  t -  /2 ) So we find  = -  /2 !! x(t) = Asin(  t) v(t) =  Acos(  t) a(t) = -  2 Asin(  t)  tt x(t)A -A

21. Physics 1501: Lecture 25, Pg 21 Lecture 25, Act 3 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 (a) v(t) = - v max sin(  t) a(t) = -a max cos(  t) (b) v(t) = v max sin(  t) a(t) = a max cos(  t) (c) v(t) = v max cos(  t) a(t) = -a max cos(  t) (both v max and a max are positive numbers) t = 0

22. Physics 1501: Lecture 25, Pg 22 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 +  ) Remember,

23. Physics 1501: Lecture 25, Pg 23 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    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

24. Physics 1501: Lecture 25, Pg 24 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!

25. Physics 1501: Lecture 25, Pg 25 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

26. Physics 1501: Lecture 25, Pg 26 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   L d m mg z where Differential equation for simple harmonic motion !  =  0 cos(  t +  ) But  = I  I  =  mL 2

27. Physics 1501: Lecture 25, Pg 27 Lecture 25, Act 4 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

28. Physics 1501: Lecture 25, Pg 28 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

29. Physics 1501: Lecture 25, Pg 29 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 d I So  = I  becomes

30. Physics 1501: Lecture 25, Pg 30 Lecture 25, Act 25 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