1. Chapter 12 Oscillations

2. 2 Mechanical oscillations: pendulum, string of a guitar, vocal cords, … More general oscillations: Electrical, optical, atomic, … Oscillations: motions that repeat themselves An object moves back and forth repeatedly around the equilibrium position. Or a physical quantity changes repeatedly around a fixed value.

3. Analysis of oscillations 3 Simple harmonic motion (SHM) is the most simple and basic oscillation. Every oscillation can be composed of SHM.

4. Oscillations of a spring 4 Horizontal Spring-mass system 1-D motion, origin o is the equilibrium position x is the displacement of the object relative to o F is called restoring force x > 0: stretched, F < 0 x 0 ( Hooke’s law )

5. Dynamic equation of SHM 5 Any vibrating system with a restoring force exhibits a simple harmonic motion (SHM) where m is the mass which is oscillating This is called the dynamic equation of SHM angular frequency

6. Vibrating cube 6 Example1: A woody cube can float with height h under water. If it is pushed down and released, there will be a SHM. Determine the angular frequency. Solution: Assume a mass m and length l h o x x With displacement x

7. Thinking question 7 Thinking: A mass M is connected to two springs (k 1, k 2 ). What is the angular frequency? M k1k1 k2k2 F x x1x1 x2x2

8. Motional equation of SHM 8 motional equation 1) Amplitude A : the maximum magnitude of displacement from equilibrium 2) Angular Frequency ω : rate of the vibration determined by the system A and : integral constants → initial conditions

9. Quantities in SHM 9 Period T : time required for one complete cycle Frequency f : number of complete cycles per sec Phase: key point in oscillations or waves natural frequency 3) : the phase of SHM at time t, note as θ : phase at t=0, initial phase or phase angle

10. Phase difference 10 1) For one SHM at different time 2) For two SHM with same ω at same time >0: x 2 is in front of x 1 =0: x 1 and x 2 are in phase <0: x 2 is behind of x 1 =π: x 1 and x 2 are inverse phase

11. Velocity and acceleration 11 Can also be treated as SHM. Phase difference?

12. Determine motional equation 12 determined by the system A and : integral constants → initial conditions When t=0:

13. Spring oscillator 13 Example2: A 4kg mass is attached to a spring of k=100N/m. It has an initial velocity v 0 = - 5m/s and initial displacement x 0 =1.0m. Motional equation? Solution: Angular frequency Amplitude Phase angle

14. Rotational vector method 14 SHMUniform circular motion (  ) (-  /2)  M A o x  t+φ p x (0) (+  /2) OM rotates with constant  counterclockwise OM =A p is the projection of M on x axis, the position of P: This is a SHM! Angle  t+φ is the phase Rotational vector

15. Geometric description 15 UCMtwo SHMs in x and y direction SHMUCM AAmplitudeRadius  Angular FrequencyAngular Velocity xDisplacementProjection on x-axis  PhaseAngle makes with +x x Show rotational vectors 1) x= - A,  t+φ =  2) x=0, v>0: 3) x=A/2, v<0:  t+φ = -  /2  t+φ =  /3

16. Rotational vector in SHM 16 Example3: T=2s, A=0.12m. When t=0, x 0 =0.06m, the object moves in positive direction. Determine: a) motional equation Solution: a) b) the phase when x= - 0.06m, v<0 x c) minimum time required from position b to E.P. rotation angle

17. Graph of function 17 Example4: Determine the motional function. Solution: A = 6 cm 2 x (cm) o t (s) 6 3 ( )cm o  x  /3 t=0, t=2 s

18. Energy in SHM 18 Kinetic energy Potential energy where Total mechanical energy is conserved in SHM Total mechanical energy

19. Pushing on a spring 19 Example5: No friction, k=16N/m, m=4kg. A force F=10N pushes m moving s=5cm, and we let t=0 at the time when m reaches the far left. Determine: a) motional function Solution: a) s m F k x o Vibration energy is from work done by F b) at what position E k = E p ?

20. Massive spring system 20 Example6: A massive spring m s (k, L), attached by a mass M, determine the angular frequency. (no fr) Ox l dl M Solution: We have already obtained ( Ch7, P15 ) Conservation of total mechanical energy

21. Physical pendulum 21 Extended body oscillates in vertical plane For a small angle  h o C mg 

22. Simple pendulum and more 22 Example7: Determine the period of pendulums: a) b) c)  l  l  m, l M, R Solution: a) Simple pendulum b) c) ? Measure of INatural walking pace

23. Longest tunnel 23 Thinking: If there is a tunnel that extends from one side of the earth, through its center, to the other side. Determine the time required for a travel through it. Solution:

24. Superposition of SHM 24 Superposition: the sum of several motions 1. Same Direction, Same Frequency SHM with same frequency O x 11 A1A1 22 A2A2  ω A  can also be known, but it is not important

25. Constructive & destructive superposition 25 1) in phaseconstructive superposition 2) inverse phasedestructive superposition

26. General superposition 26 Example8: Determine the superposition of x 1 and x 2 : x 1 =3cos(  t+  /2) ； x 2 =3cos(  t+  ) Solution: x

27. Homework 27 The superposition of x 1 and x 2 makes a new SHM x with amplitude A=17.3cm, where A 1 =10cm, and, determine amplitude A 2.

28. Different frequency 28 2. Same Direction, Different Frequency For a special case: Amplitude varying oscillation —— Beats where Amplitude modulation factor

29. Amplitude modulation 29 Amplitude modulation: AM Frequency modulation: FM x1x1 x2x2 x

30. *Superposition in 2D 30 3. Vertical Direction, Same Frequency Motion in an ellipse ** Special case: rotational vector method 4. Vertical Direction, Different Frequency 1 : 1 1 : 2 4 : 3 9 : 74 : 3 Lissajous Figures

31. Damped harmonic motion 31 Amplitude of real oscillator slowly decreases Damping force depends on the speed F = F(v) Dynamic equation It is called damped harmonic motion (a simple case)

32. Solution of Damped motion 32 where 1) Underdamped motion: ① frequency becomes lower mean lifetime ② amplitude keeps decreasing over time

33. 33 2) Overdamped motion: No longer an oscillation! 3) Critical damped motion: No oscillation Car’s spring …

34. Forced vibrations 34 DampedExternal force to supply energy Natural frequency Driving force has its own frequency f The amplitude becomes large when f is near f 0 f = f 0 → maximum vibration energy Resonance f 0 : resonance frequency and f is the frequency of vibration!

35. *Resonance 35 Energy of driving force is efficiently transformed Resonance causes the collapse of bridge * Nuclear magnetic resonance (NMR) Selection of television channel