1. Physics 1501: Lecture 26, Pg 1 Physics 1501: Lecture 26 Today’s Agenda l Homework #9 (due Friday Nov. 4) l Midterm 2: Nov. 16 l Katzenstein Lecture: Nobel Laureate Gerhard t’Hooft çFriday at 4:00 in P-36 … l Topics çSimple Harmonic Motion – masses on springs çPendulum çEnergy of the SHO

2. Physics 1501: Lecture 26, Pg 2 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

3. Physics 1501: Lecture 26, Pg 3 SHM So Far l We showed that (which came from F=ma) has the most general solution 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!

4. Physics 1501: Lecture 26, Pg 4 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

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

6. Physics 1501: Lecture 26, Pg 6 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

7. Physics 1501: Lecture 26, Pg 7 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

8. Physics 1501: Lecture 26, Pg 8 Lecture 26, Act 1 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

9. Physics 1501: Lecture 26, Pg 9 Suppose we have some arbitrarily shaped solid of mass M hung on a fixed axis, that we know where the CM is located and what the moment of inertia I about the axis is. The torque about the rotation (z) axis for small  is (sin    )  = - Mgd  - MgR  General Physical Pendulum  d Mg z-axis R x CM where  =  0 cos(  t +  )  

10. Physics 1501: Lecture 26, Pg 10 Lecture 26, Act 2 Physical Pendulum l A pendulum is made by hanging a thin hoola-hoop of diameter D on a small nail.  What is the angular frequency of oscillation of the hoop for small displacements ? ( I CM = mR 2 for a hoop) (a) (b) (c) D pivot (nail)

11. Physics 1501: Lecture 26, Pg 11 Torsion Pendulum Consider an object suspended by a wire attached at its CM. The wire defines the rotation axis, and the moment of inertia I about this axis is known. l The wire acts like a “rotational spring”. çWhen the object is rotated, the wire is twisted. This produces a torque that opposes the rotation.  In analogy with a spring, the torque produced is proportional to the displacement:  = -k  I wire  

12. Physics 1501: Lecture 26, Pg 12 Torsion Pendulum... Since  = -k   = I  becomes I wire   Similar to “mass on spring”, except I has taken the place of m (no surprise) where

13. Physics 1501: Lecture 26, Pg 13 Lecture 26, Act 3 Period l All of the following pendulum bobs have the same mass. Which pendulum rotates the fastest, i.e. has the smallest period? (The wires are identical) RRRR A) B) C) D)

14. Physics 1501: Lecture 26, Pg 14 l Spring-mass system l Pendula çGeneral physical pendulum »Simple pendulum çTorsion pendulum Simple Harmonic Oscillator  d Mg z-axis R x CM  =  0 cos(  t +  ) k x m F F = -kxa I wire   x(t) = Acos(  t +  ) where

15. Physics 1501: Lecture 26, Pg 15 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 Active Figure

16. Physics 1501: Lecture 26, Pg 16 Energy in SHM l For both the spring and the pendulum, we can derive the SHM solution using energy conservation. l The total energy (K + U) of a system undergoing SMH will always be constant! l This is not surprising since there are only conservative forces present, hence energy is conserved. -AA0 s U U K E

17. Physics 1501: Lecture 26, Pg 17 SHM and quadratic potentials l SHM will occur whenever the potential is quadratic. l Generally, this will not be the case: l For example, the potential between H atoms in an H 2 molecule looks something like this: -AA0 x U U K E U x

18. Physics 1501: Lecture 26, Pg 18 SHM and quadratic potentials... However, if we do a Taylor expansion of this function about the minimum, we find that for small displacements, the potential IS quadratic: U x U(x) = U(x 0 ) + U(x 0 ) (x- x 0 ) + U  (x 0 ) (x- x 0 ) 2 +.... U(x) = 0 (since x 0 is minimum of potential) x0x0 U x Define x = x - x 0 and U(x 0 ) = 0 Then U(x) = U  (x 0 ) x 2

19. Physics 1501: Lecture 26, Pg 19 SHM and quadratic potentials... U x x0x0 U x U(x) = U  (x 0 ) x 2 Let k = U  (x 0 ) Then: U(x) = k x 2 SHM potential !!

20. Physics 1501: Lecture 26, Pg 20 What about Friction? l Friction causes the oscillations to get smaller over time l This is known as DAMPING. l As a model, we assume that the force due to friction is proportional to the velocity.

21. Physics 1501: Lecture 26, Pg 21 What about Friction? We can guess at a new solution. With,

22. Physics 1501: Lecture 26, Pg 22 What about Friction? What does this function look like? (You saw it in lab, it really works)

23. Physics 1501: Lecture 26, Pg 23 What about Friction? There is a cuter way to write this function if you remember that exp(ix) = cos x + i sin x.

24. Physics 1501: Lecture 26, Pg 24 Damped Simple Harmonic Motion l Frequency is now a complex number! What gives? çReal part is the new (reduced) angular frequency çImaginary part is exponential decay constant underdamped critically damped overdamped Active Figure

25. Physics 1501: Lecture 26, Pg 25 Driven SHM with Resistance l To replace the energy lost to friction, we can drive the motion with a periodic force. (Examples soon). l Adding this to our equation from last time gives, F = F 0 cos(  t)

26. Physics 1501: Lecture 26, Pg 26 Driven SHM with Resistance l So we have the equation, l As before we use the same general form of solution, l Now we plug this into the above equation, do the derivatives, and we find that the solution works as long as,

27. Physics 1501: Lecture 26, Pg 27 Driven SHM with Resistance l So this is what we need to think about, I.e. the amplitude of the oscillating motion, l Note, that A gets bigger if F o does, and gets smaller if b or m gets bigger. No surprise there. l Then at least one of the terms in the denominator vanishes and the amplitude gets real big. This is known as resonance. l Something more surprising happens if you drive the pendulum at exactly the frequency it wants to go,

28. Physics 1501: Lecture 26, Pg 28 Driven SHM with Resistance l Now, consider what b does,   b small b middling b large  

29. Physics 1501: Lecture 26, Pg 29 Dramatic example of resonance l In 1940, turbulent winds set up a torsional vibration in the Tacoma Narrow Bridge 

30. Physics 1501: Lecture 26, Pg 30 Dramatic example of resonance  l when it reached the natural frequency

31. Physics 1501: Lecture 26, Pg 31 Dramatic example of resonance  l it collapsed ! Other example: London Millenium Bridge

32. Physics 1501: Lecture 26, Pg 32 Lecture 26, Act 4 Resonant Motion l Consider the following set of pendula all attached to the same string D A B C If I start bob D swinging which of the others will have the largest swing amplitude ? (A)(B)(C)