Physics 111: Lecture 24 Today’s Agenda

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

2. k m k m k m Simple Harmonic Motion (SHM) Horizontal Spring • 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...

3. F = -kx a k m x SHM Dynamics • At any given instant we know thatF = mamust be true. • But in this case F = -kx and ma = • So: -kx = ma = a differential equation for x(t)!

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

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

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. = C cos(t)+ B sin(t) where C = A cos() and B = A sin() It works! Derivation: We want to use the most general solution: 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 So we can use x = A cos(t + )as the most general solution!

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

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

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

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

12. Lecture 24, Act 1Solution • 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

13. m What about Vertical Springs? Vertical Spring • 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 y = 0 F = -ky Which has solution y = A cos(t + ) where

14. SHM So Far • The most general solution is x = A cos(t + ) where A = amplitude  = 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 force is zero!

15. z  L m mg The Simple Pendulum Simple Pendulum • 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.

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

17. z  L where m Differential equation for simple harmonic motion! d  = 0 cos(t + ) mg 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

18. Lecture 24, Act 2Simple 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

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

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

21. The Rod Pendulum • 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. z  x CM L mg

22. where 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 z I d L/2  x CM L d mg

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

24. LS LR S = Pif Lecture 24, Act 3Solution

25. 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: # 1, 13, 33, 55, 93