1. Physics 111: Lecture 24, Pg 1 Physics 111: Lecture 24 Today’s Agenda l Introduction to Simple Harmonic Motion çHorizontal spring & mass l The meaning of all these sines and cosines l Vertical spring & mass l The energy approach l The simple pendulum l The rod pendulum

2. Physics 111: Lecture 24, 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 Horizontal Spring

3. Physics 111: Lecture 24, 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)!

4. Physics 111: Lecture 24, Pg 4 SHM Dynamics... Try the solution x = A cos(  t) This works, so it must be a solution! define Where  is the angular frequency of motion

5. Physics 111: Lecture 24, 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 ?? Movie (shm) x y 1 0  1 1 22 33 4 4 55 66 Shadow

6. Physics 111: Lecture 24, Pg 6 SHM Solution We just showed that (which came from F = ma) has the solution x = A cos(  t). This is not a unique solution, though. x = A sin(  t) is also a solution. The most general solution is a linear combination of these two solutions! x = B sin(  t)+ C cos(  t) ok

7. Physics 111: Lecture 24, Pg 7 Derivation: x = A cos(  t +  ) is equivalent to x = B sin(  t)+ C cos(  t) x = A cos(  t +  ) = A cos(  t) cos  - A sin(  t) sin  where C = A cos(  ) and B =  A sin(  ) It works! = C cos(  t) + B sin(  t) We want to use the most general solution: So we can use x = A cos(  t +  ) as the most general solution!

8. Physics 111: Lecture 24, Pg 8 SHM Solution... Drawing of A cos(  t ) l A = amplitude of oscillation    T = 2  /  A A 

9. Physics 111: Lecture 24, Pg 9 SHM Solution... Drawing of A cos(  t +  )     

10. Physics 111: Lecture 24, Pg 10 SHM Solution... Drawing of A cos(  t -  /2) A  =  /2    = A sin(  t)! 

11. Physics 111: Lecture 24, Pg 11 Lecture 24, Act 1 Simple Harmonic Motion l If you added the two sinusoidal waves shown in the top plot, what would the result look like? (a) (b) (c)

12. Physics 111: Lecture 24, Pg 12 Lecture 24, Act 1 Solution l Recall your trig identities: So Where l The sum of two or more sines or cosines having the same frequency is just another sine or cosine with the same frequency. l The answer is (b). Prove this with Excel

13. Physics 111: Lecture 24, Pg 13 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 = A cos(  t +  ) where Vertical Spring

14. Physics 111: Lecture 24, Pg 14 SHM So Far The most general solution is x = A cos(  t +  ) where A = amplitude  = frequency  = phase 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!

15. Physics 111: Lecture 24, Pg 15 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 Simple Pendulum

16. Physics 111: Lecture 24, Pg 16 Aside: sin  and cos  for small  l A Taylor expansion of sin  and cos  about  = 0 gives: and So for  << 1, and

17. Physics 111: Lecture 24, Pg 17 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 +  )

18. Physics 111: Lecture 24, Pg 18 Lecture 24, Act 2 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

19. Physics 111: Lecture 24, Pg 19 Lecture 24, Act 2 Solution l We have shown that for a simple pendulum Since l If we make a pendulum shorter, it oscillates faster (smaller period)

20. Physics 111: Lecture 24, Pg 20 Lecture 24, Act 2 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.

21. Physics 111: Lecture 24, Pg 21 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

22. Physics 111: Lecture 24, Pg 22 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 So  = I  becomes  L d mg z L/2 x CM where d I

23. Physics 111: Lecture 24, Pg 23 Lecture 24, Act 3 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 Physical Pendulum

24. Physics 111: Lecture 24, Pg 24 LRLR LSLS  S =  P if Lecture 24, Act 3 Solution

25. Physics 111: Lecture 24, Pg 25 Recap of today’s lecture l Introduction to Simple Harmonic Motion (Text: 14-1) çHorizontal spring & mass l The meaning of all these sines and cosines l Vertical spring & mass(Text: 14-3) l The energy approach(Text: 14-2) l The simple pendulum (Text: 14-3) l The rod pendulum l Look at textbook problems l Look at textbook problems Chapter 14: # 1, 13, 33, 55, 93