1. Copyright © 2010 Pearson Education, Inc. Lecture Outline Chapter 13 Physics, 4 th Edition James S. Walker

2. Copyright © 2010 Pearson Education, Inc. Chapter 13 Oscillations about Equilibrium

3. Copyright © 2010 Pearson Education, Inc. Units of Chapter 13 Periodic Motion Simple Harmonic Motion Connections between Uniform Circular Motion and Simple Harmonic Motion The Period of a Mass on a Spring Energy Conservation in Oscillatory Motion

4. Copyright © 2010 Pearson Education, Inc. Units of Chapter 13 The Pendulum Damped Oscillations Driven Oscillations and Resonance

5. Copyright © 2010 Pearson Education, Inc. 13-1 Periodic Motion Period: time required for one cycle of periodic motion Frequency: number of oscillations per unit time This unit is called the Hertz:

6. Copyright © 2010 Pearson Education, Inc. 13-2 Simple Harmonic Motion A spring exerts a restoring force that is proportional to the displacement from equilibrium: A mass attached to a spring undergoes simple harmonic motion about x = 0 (a) The mass is at its maximum positive value of x. Its velocity is zero, and the force on it points to the left with maximum magnitude. (b) The mass is at the equilibrium position of the spring. Here the speed has its maximum has its maximum value, and the force exerted by the spring is zero. (c) The mass is at its maximum displacement in the negative x direction. The velocity is zero is zero here, and the force points to the right with maximum magnitude. (d) The mass is at the equilibrium position of the spring with zero force acting on it and maximum speed. (e) The mass has completed one cycle of its oscillation about x = 0.

7. Copyright © 2010 Pearson Education, Inc. 13-2 Simple Harmonic Motion A mass on a spring has a displacement as a function of time that is a sine or cosine curve: Here, A is called the amplitude of the motion. Displaying position versus time for simple harmonic motion As an air-track cart oscillates about its equilibrium position, a pen attached to it traces its motion onto a moving sheet of paper. This produces a “strip chart”, showing that the cart’s motion has the shape of a sine or a cosine.

8. Copyright © 2010 Pearson Education, Inc. 13-2 Simple Harmonic Motion If we call the period of the motion T – this is the time to complete one full cycle – we can write the position as a function of time:

9. Copyright © 2010 Pearson Education, Inc. 13-2 Simple Harmonic Motion An air-track cart attached to a spring completes one oscillation every 2.4 s. At t=0, the cart is released from rest at a distance of 0.10 m from its equilibrium position. What is the position of the cart at (a) 0.30 s, (b) 0.6 s ( c) 2.7 s X = (0.10 m) cos [( 2π/2.4 s) (0.30 s)] = (0.10 m) cos (π /4) = 7.1 cm X = (0.10 m) cos [( 2π/2.4 s) (0.60 s)] = (0.10 m) cos (π /2) = 0 X = (0.10 m) cos [( 2π/2.4 s) (2.7 s)] = (0.10 m) cos (9π /4) = 7.1 cm

10. Copyright © 2010 Pearson Education, Inc. 13-3 Connections between Uniform Circular Motion and Simple Harmonic Motion An object in simple harmonic motion has the same motion as one component of an object in uniform circular motion:

11. Copyright © 2010 Pearson Education, Inc. 13-3 Connections between Uniform Circular Motion and Simple Harmonic Motion Here, the object in circular motion has an angular speed of where T is the period of motion of the object in simple harmonic motion. Θ = ω t That is the angular position increases linearly with time

12. Copyright © 2010 Pearson Education, Inc. 13-3 Connections between Uniform Circular Motion and Simple Harmonic Motion The position as a function of time: The angular frequency, ω:

13. Copyright © 2010 Pearson Education, Inc. 13-3 Connections between Uniform Circular Motion and Simple Harmonic Motion The velocity as a function of time: And the acceleration: Si unit: m/s 2 Both of these are found by taking components of the circular motion quantities.

14. Copyright © 2010 Pearson Education, Inc. 13-3 Connections between Uniform Circular Motion and Simple Harmonic Motion As in example 13-1, an air-track cart attached to a spring completes one oscillation every 2.4 s. At t =0 the cart is released from rest with the spring stretched 0.10 m from its equilibrium position. What are the velocity and acceleration of the cart at (a) 0.30 s and (b) 0.60 s? ω = 2π/T = 2 π/ (2.4 s) = 2.6 rad /s V = - (0.10 m) (2.6 rad /s) sin [ ( 2π / 2.4 s) (0.30 s)] V = - (26 cm/s) sin (π /4) = - 18 cm/s a = - (0.10 m) (2.6 rad /s) 2 cos [ ( 2π / 2.4 s) (0.30 s)] = - (68 cm / s 2 ) cos (π /4) = - 48 cm /s 2

15. Copyright © 2010 Pearson Education, Inc. 13-4 The Period of a Mass on a Spring Since the force on a mass on a spring is proportional to the displacement, and also to the acceleration, we find that. Substituting the time dependencies of a and x gives

16. Copyright © 2010 Pearson Education, Inc. 13-4 The Period of a Mass on a Spring Therefore, the period is

17. Copyright © 2010 Pearson Education, Inc. 13-4 The Period of a Mass on a Spring On December 29,1997, a United Airlines flight from Tokyo to Honolulu was hit with severe turbulence 31 minutes after takeoff. Data from the airplane’s “black box” indicated the 747 moved up and down with an amplitude of 30.0 m and a maximum acceleration of 1.8g. Treating the up- and –down motion of the plane as simple harmonic, find (a) the time required for one complete oscillation and (b) the plane’s maximum vertical speed. ( c) what amplitude of motion would result in a maximum acceleration of 0.50g, everything remaining the same?

18. Copyright © 2010 Pearson Education, Inc. 13-4 The Period of a Mass on a Spring A 0.120 kg mass attached to a spring oscillates with an amplitude of 0.0750 m and a maximum speed of 0.524 m/s. Find (a) the force constant and (b) the period of motion.

19. Copyright © 2010 Pearson Education, Inc. 13-4 The Period of a Mass on a Spring Now, when a mass m is attached to a vertical spring, it causes the spring to stretch. In fact, the vertical spring is in equilibrium when it exerts an upward force equal to the weight of the mass. That is, the spring stretches by an amount y o given by ky o = mg y o = mg / k.

20. Copyright © 2010 Pearson Education, Inc. 13-4 The Period of a Mass on a Spring A 0.260 kg mass is attached to a vertical spring. When the mass is put into motion, its period is 1.12 s. (a) How much does the mass stretch the spring when it is at rest in its equilibrium position? (b) Suppose this experiment is repeated on a planet where the acceleration due to gravity is twice what it is on Earth. By what multiplicative factors do the period and equilibrium stretch change? T = 2 π (m / k) 1/2 K = 4π 2 m / T 2 = 4 π 2 (0.260 kg) / 1.12 s) 2 = 8.18 kg / s 2 K = 8.18 N/m Ky o = mg y o = mg / k = (0.26 kg)(9.81 m/s 2 ) / 8.18 N/m y o = 0.312 m (b) T = 2 π (m / k) 1/2 g -  2g y o = mg / k -  2y o

21. Copyright © 2010 Pearson Education, Inc. 13-5 Energy Conservation in Oscillatory Motion In an ideal system with no nonconservative forces, the total mechanical energy is conserved. For a mass on a spring: Since we know the position and velocity as functions of time, we can find the maximum kinetic and potential energies:

22. Copyright © 2010 Pearson Education, Inc. 13-5 Energy Conservation in Oscillatory Motion As a function of time, So the total energy is constant; as the kinetic energy increases, the potential energy decreases, and vice versa.

23. Copyright © 2010 Pearson Education, Inc. 13-5 Energy Conservation in Oscillatory Motion This diagram shows how the energy transforms from potential to kinetic and back, while the total energy remains the same.

24. Copyright © 2010 Pearson Education, Inc. 13-5 Energy Conservation in Oscillatory Motion A 0.98 kg block slides on a frictionless, horizontal surface with a speed of 1.32 m/s. The block encounters an unstretched spring with a force constant of 245 N /m, as shown in the sketch. (a) How far is the spring compressed before the block comes to rest? (b) How long is the block in contact with the spring before it comes to rest? 1/2mv o = 1/2KA 2 A = V o (m/k) 1/2 = (1.32 m /s )(0.98 kg) (245 N / m) 1/2 A = 0.0835 m Calculate the period of one oscillation T = 2π (m/k) 1/2 = 2π (0.98 kg/ 245 N/m) 1/2 = 0.397 s In moving from the equilibrium position of the spring to maximum compression, the mass has undergone one-quarter of a cycle; thus the time is T/4. t = 1/4T = ¼ (0.397 s) = 0.0993 s

25. Copyright © 2010 Pearson Education, Inc. 13-6 The Pendulum A simple pendulum consists of a mass m (of negligible size) suspended by a string or rod of length L (and negligible mass). The angle it makes with the vertical varies with time as a sine or cosine.

26. Copyright © 2010 Pearson Education, Inc. 13-6 The Pendulum Looking at the forces on the pendulum bob, we see that the restoring force is proportional to sin θ, whereas the restoring force for a spring is proportional to the displacement (which is θ in this case).

27. Copyright © 2010 Pearson Education, Inc. 13-6 The Pendulum However, for small angles, sin θ and θ are approximately equal. Relationship between sin θ and θ. For small angles measured in radians, sin θ is approximately equal to θ. Thus, when considering small oscillations of a pendulum, we can replace sin θ with θ. Period of a Pendulum (small amplitude) T = 2 π (L / g) ½ SI unit : s

28. Copyright © 2010 Pearson Education, Inc. 13-6 The Pendulum Substituting θ for sin θ allows us to treat the pendulum in a mathematically identical way to the mass on a spring. Therefore, we find that the period of a pendulum depends only on the length of the string:

29. Copyright © 2010 Pearson Education, Inc. 13-6 The Pendulum A pendulum is constructed from a string 0.627 m long attached to a mass of 0.250 kg. When set in motion, the pendulum completes one oscillation every 1.59 s. If the pendulum is held at rest and the string is cut, how long will it take for the mass to fall through a distance of 1.0 m? T g = 4 π 2 L / T2 g = 4 π 2 (0.627 m) / ( 1.59 s) 2 = 9.79 m/s 2 y = 1/2gt 2 or t = (2y / g ) ½ t = (2(1.0 m) / 9.79 m / s 2 ) 1/2 = 0.452 s

30. Copyright © 2010 Pearson Education, Inc. 13-6 The Pendulum A physical pendulum is a solid mass that oscillates around its center of mass, but cannot be modeled as a point mass suspended by a massless string. Examples: Examples of physical pendula In each case, an object of definite size and shape oscillates about a given pivot point. The period of oscillation depends in detail on the location of the pivot point as well as on the distance l from it to the center of mass, CM.

31. Copyright © 2010 Pearson Education, Inc. 13-6 The Pendulum In this case, it can be shown that the period depends on the moment of inertia: Substituting the moment of inertia of a point mass a distance l from the axis of rotation gives, as expected,

32. Copyright © 2010 Pearson Education, Inc. 13-7 Damped Oscillations In most physical situations, there is a nonconservative force of some sort, which will tend to decrease the amplitude of the oscillation, and which is typically proportional to the speed: This causes the amplitude to decrease exponentially with time:

33. Copyright © 2010 Pearson Education, Inc. 13-7 Damped Oscillations This exponential decrease is shown in the figure:

34. Copyright © 2010 Pearson Education, Inc. 13-7 Damped Oscillations The previous image shows a system that is underdamped – it goes through multiple oscillations before coming to rest. A critically damped system is one that relaxes back to the equilibrium position without oscillating and in minimum time; an overdamped system will also not oscillate but is damped so heavily that it takes longer to reach equilibrium.

35. Copyright © 2010 Pearson Education, Inc. 13-8 Driven Oscillations and Resonance An oscillation can be driven by an oscillating driving force; the frequency of the driving force may or may not be the same as the natural frequency of the system.

36. Copyright © 2010 Pearson Education, Inc. 13-8 Driven Oscillations and Resonance If the driving frequency is close to the natural frequency, the amplitude can become quite large, especially if the damping is small. This is called resonance.

37. Copyright © 2010 Pearson Education, Inc. Summary of Chapter 13 Period: time required for a motion to go through a complete cycle Frequency: number of oscillations per unit time Angular frequency: Simple harmonic motion occurs when the restoring force is proportional to the displacement from equilibrium.

38. Copyright © 2010 Pearson Education, Inc. Summary of Chapter 13 The amplitude is the maximum displacement from equilibrium. Position as a function of time: Velocity as a function of time:

39. Copyright © 2010 Pearson Education, Inc. Summary of Chapter 13 Acceleration as a function of time: Period of a mass on a spring: Total energy in simple harmonic motion:

40. Copyright © 2010 Pearson Education, Inc. Summary of Chapter 13 Potential energy as a function of time: Kinetic energy as a function of time: A simple pendulum with small amplitude exhibits simple harmonic motion

41. Copyright © 2010 Pearson Education, Inc. Summary of Chapter 13 Period of a simple pendulum: Period of a physical pendulum:

42. Copyright © 2010 Pearson Education, Inc. Summary of Chapter 13 Oscillations where there is a nonconservative force are called damped. Underdamped: the amplitude decreases exponentially with time: Critically damped: no oscillations; system relaxes back to equilibrium in minimum time Overdamped: also no oscillations, but slower than critical damping

43. Copyright © 2010 Pearson Education, Inc. Summary of Chapter 13 An oscillating system may be driven by an external force This force may replace energy lost to friction, or may cause the amplitude to increase greatly at resonance Resonance occurs when the driving frequency is equal to the natural frequency of the system