Physics 211: Lecture 24Today’s Agenda: Physics 211: Lecture 24 Today’s Agenda Introduction to Simple Harmonic Motion Horizontal spring & mass (no gravity) The meaning of all these sines and cosines they are solutions to the differential equation Vertical spring & mass (gravity) The energy approach The simple pendulum The rod pendulum


Simple Harmonic Motion (SHM): 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... Horizontal Spring


SHM Dynamics: SHM Dynamics At any given instant we know that F = ma must be true. But in this case F = -kx and ma = So: -kx = ma = Our job is to find x(t)


SHM Dynamics...: SHM Dynamics... define


SHM Dynamics...: SHM Dynamics... y = R cos  = R cos (t) But wait a minute...what does angular frequency  have to do with moving back & forth in a straight line ?? Movie (shm) x y -1 1 0  1 1 2 2 3 3 4 4 5 5 6 6 Shadow


SHM Solution: 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


Derivation:: 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 We want to use the most general solution: So we can use x = A cos(t + ) as the most general solution! Still two parameters to specify, amplitude and phase…


SHM Solution...: SHM Solution... Drawing of A cos(t ) A = amplitude of oscillation     T = 2/ A A -


SHM Solution...: SHM Solution... Drawing of A cos(t + )      -


SHM Solution...: SHM Solution... Drawing of A cos(t - /2) = A sin(wt) A  = /2     = A sin(t)! -


Meaning of the solution to SHM Solution...: Meaning of the solution to SHM Solution... Drawing of A cos(t + ) What does this really mean? Just that the amplitude and phase are both able to be set


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


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


What about Vertical Springs?: What about Vertical Springs? We already know that for a vertical spring if y is measured from the equilibrium position The force of the spring is the negative derivative of this function: So this will be just like the horizontal case: -ky = ma = j k F = -ky y = 0 Which has solution y = A cos(t + ) where Vertical Spring


Slide15: U of Spring + Gravity -60 -40 -20 0 20 40 60 80 100 120 140 160 -10 -8 -6 -4 -2 0 2 4 6 8 10 y U UG = mgy US = 1/2ky2 UNET = UG + US shift due to mgy term 0 ye


What about Vertical Springs?Alternate treatment: What about Vertical Springs? Alternate treatment


SHM So Far: SHM So Far The most general solution is x = A cos(t + ) where A = amplitude  = angular frequency  = phase 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 net force is zero!


The Simple Pendulum: The Simple Pendulum A pendulum is made by suspending a mass m at the end of a string of length L. Find the angular frequency of oscillation for small displacements. Simple Pendulum


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


The Simple Pendulum...: 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


Lecture 24, Act 2Simple Harmonic Motion: Lecture 24, Act 2 Simple 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


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


Lecture 24, Act 2 Solution: Lecture 24, Act 2 Solution L1 L2 Standing up raises the CM of the swing, making it shorter! T1 T2 Since L1 > L2 we see that T1 > T2 .


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


The Rod Pendulum...: 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 So  = Ibecomes  L d mg z L/2 x CM d I


Lecture 24, Act 3Period: Lecture 24, Act 3 Period (a) (b) (c) What length do we make the simple pendulum so that it has the same period as the rod pendulum? Physical Pendulum


Lecture 24, Act 3Solution: Lecture 24, Act 3 Solution


Recap of today’s lecture: Recap of today’s lecture Introduction to Simple Harmonic Motion (Text: 14-1) Horizontal spring & mass The meaning of all these sines and cosines Vertical spring & mass (Text: 14-3) The energy approach (Text: 14-2) The simple pendulum (Text: 14-3) The rod pendulum Look at textbook problems Chapter 14: # 29, 45, 65, 93