1. Simple Harmonic Motion

2. l Vibrations è Vocal cords when singing/speaking è String/rubber band l Simple Harmonic Motion è Restoring force proportional to displacement è Springs F = -kx 11

3. Question II A mass on a spring oscillates back & forth with simple harmonic motion of amplitude A. A plot of displacement (x) versus time (t) is shown below. At what points during its oscillation is the magnitude of the acceleration of the block biggest? 1. When x = +A or -A (i.e. maximum displacement) 2. When x = 0 (i.e. zero displacement) 3. The acceleration of the mass is constant +A t -A x 17

4. Potential Energy in Spring l Force of spring is Conservative è F = -k x è W = -1/2 k x 2 è Work done only depends on initial and final position è Define Potential Energy PE spring = ½ k x 2 Force x work 20

5. ***Energy in SHM*** l A mass is attached to a spring and set to motion. The maximum displacement is x=A è Energy = PE + KE = constant! = ½ k x 2 + ½ m v 2 è At maximum displacement x=A, v = 0 Energy = ½ k A 2 + 0 è At zero displacement x = 0 Energy = 0 + ½ mv m 2 Since Total Energy is same ½ k A 2 = ½ m v m 2 v m = sqrt(k/m) A m x x=0 0 x PE S 25

6. Question 3 A mass on a spring oscillates back & forth with simple harmonic motion of amplitude A. A plot of displacement (x) versus time (t) is shown below. At what points during its oscillation is the speed of the block biggest? 1. When x = +A or -A (i.e. maximum displacement) 2. When x = 0 (i.e. zero displacement) 3. The speed of the mass is constant +A t -A x 29

7. Question 4 l A spring oscillates back and forth on a frictionless horizontal surface. A camera takes pictures of the position every 1/10 th of a second. Which plot best shows the positions of the mass. EndPoint Equilibrium EndPoint Equilibrium EndPoint Equilibrium 38 123123

8. X=0 X=A X=-A X=A; v=0; a=-a max X=0; v=-v max ; a=0X=-A; v=0; a=a max X=0; v=v max ; a=0 X=A; v=0; a=-a max Springs and Simple Harmonic Motion 32

9. What does moving in a circle have to do with moving back & forth in a straight line ?? y x -R R 0  1 1 22 33 4 4 55 66 R  8 7 8 7 x x = R cos  = R cos (  t ) since  =  t 34

10. SHM and Circles

11. Simple Harmonic Motion: x(t) = [A]cos(  t) v(t) = -[A  ]sin(  t) a(t) = -[A  2 ]cos(  t) x(t) = [A]sin(  t) v(t) = [A  ]cos(  t) a(t) = -[A  2 ]sin(  t) x max = A v max = A  a max = A  2 Period = T (seconds per cycle) Frequency = f = 1/T (cycles per second) Angular frequency =  = 2  f = 2  /T For spring:  2 = k/m OR 36

12. Example A 3 kg mass is attached to a spring (k=24 N/m). It is stretched 5 cm. At time t=0 it is released and oscillates. Which equation describes the position as a function of time x(t) = A) 5 sin(  t)B) 5 cos(  t)C) 24 sin(  t) D) 24 cos(  t)E) -24 cos(  t) 39

13. Example A 3 kg mass is attached to a spring (k=24 N/m). It is stretched 5 cm. At time t=0 it is released and oscillates. What is the total energy of the block spring system? 43

14. Example A 3 kg mass is attached to a spring (k=24 N/m). It is stretched 5 cm. At time t=0 it is released and oscillates. What is the maximum speed of the block? 46

15. Example A 3 kg mass is attached to a spring (k=24 N/m). It is stretched 5 cm. At time t=0 it is released and oscillates. How long does it take for the block to return to x=+5cm? 49

16. Pendulum Motion l For small angles è T = mg è T x = -mg (x/L) Note: F proportional to x!   F x = m a x -mg (x/L) = m a x a x = -(g/L) x  Recall for SHO a = -  2 x  = sqrt(g/L)  T = 2  sqrt(L/g)  Period does not depend on A, or m! m L x T mg 37

17. Example: Clock l If we want to make a grandfather clock so that the pendulum makes one complete cycle each sec, how long should the pendulum be?

18. Question 1 Suppose a grandfather clock (a simple pendulum) runs slow. In order to make it run on time you should: 1. Make the pendulum shorter 2. Make the pendulum longer 38

19. Summary l Simple Harmonic Motion è Occurs when have linear restoring force F= -kx  x(t) = [A] cos(  t)  v(t) = -[A  ] sin(  t)  a(t) = -[A   ] cos(  t) l Springs è F = -kx è U = ½ k x 2   = sqrt(k/m) l Pendulum (Small oscillations)   = sqrt(L/g) 50