UB, Phy101: Chapter 10, Pg 1 Physics 101: Lecture 17 Simple Harmonic Motion l Today’s lecture will cover Textbook Sections

UB, Phy101: Chapter 10, Pg 2

UB, Phy101: Chapter 10, Pg 3 Simple Harmonic Motion Period = T (seconds per cycle) Frequency = f = 1/T (cycles per second) Angular frequency =  = 2  f = 2  /T

UB, Phy101: Chapter 10, Pg 4 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

UB, Phy101: Chapter 10, Pg 5 Springs l Hooke’s Law: l Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. è F X = -k xWhere x is the displacement from the relaxed position and k is the constant of proportionality. (often called “spring constant”) relaxed position F X = 0 x x=0

UB, Phy101: Chapter 10, Pg 6 Springs l Hooke’s Law: l Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. è F X = -k xWhere x is the displacement from the relaxed position and k is the constant of proportionality. (often called “spring constant”) relaxed position F X = -kx > 0 x x  0 x=0

UB, Phy101: Chapter 10, Pg 7 Springs l Hooke’s Law: l Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. è F X = -k xWhere x is the displacement from the relaxed position and k is the constant of proportionality. (often called “spring constant”) F X = - kx < 0 x x > 0 relaxed position x=0

UB, Phy101: Chapter 10, Pg 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

UB, Phy101: Chapter 10, Pg 9 Chapter 10, Preflight 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 CORRECT

UB, Phy101: Chapter 10, Pg 10 Chapter 10, Preflight 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 CORRECT

UB, Phy101: Chapter 10, Pg 11 Review: Period of a Spring For simple harmonic oscillator  = 2  f = 2  /T For mass M on spring with spring constant k

UB, Phy101: Chapter 10, Pg 12 What does moving in a circle have to do with moving back & forth in a straight line ?? y x -R R 0  R  x x = R cos  = R cos (  t ) since  =  t

UB, Phy101: Chapter 10, Pg 13

UB, Phy101: Chapter 10, Pg 14

UB, Phy101: Chapter 10, Pg 15

UB, Phy101: Chapter 10, Pg 16 Potential Energy of a Spring 0 x Where x is measured from the equilibrium position PE S m x x=0 Analogy: marble in a bowl: PE  KE  PE  KE  PE “height” of bowl  x 2

UB, Phy101: Chapter 10, Pg 17 Same thing for a vertical spring: 0 y PE S m y Where y is measured from the equilibrium position y=0

UB, Phy101: Chapter 10, Pg 18 In either case... m y y=0 m x x=0 E total = 1/2 Mv 2 + 1/2 kx 2 = constant KE PE KE max = 1/2Mv 2 max =1/2M  2 A 2 =1/2kA 2 PE max = 1/2kA 2 E total = 1/2kA 2

UB, Phy101: Chapter 10, Pg 19

UB, Phy101: Chapter 10, Pg 20 Chapter 10, Preflight In Case 1 a mass on a spring oscillates back and forth. In Case 2, the mass is doubled but the spring and the amplitude of the oscillation is the same as in Case 1. In which case is the maximum kinetic energy of the mass the biggest? 1. Case 1 2. Case 2 3. Same CORRECT

UB, Phy101: Chapter 10, Pg 21 Chapter 10, Preflight PE = 1 / 2 kx 2 KE = 0 x=0 x=+A x=-A x=0 x=+A x=-A same for both PE = 0 KE = KE MAX same for both

UB, Phy101: Chapter 10, Pg 22 Chapter 10, Preflight The same would be true for vertical springs... PE = 1 / 2 k y 2 Y=0

UB, Phy101: Chapter 10, Pg 23

UB, Phy101: Chapter 10, Pg 24 Pendulum Summary For “small oscillation”, period does not depend on mass amplitude DEMO

UB, Phy101: Chapter 10, Pg 25 Chapter 10, Preflight 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 CORRECT

UB, Phy101: Chapter 10, Pg 26 Chapter 10, Preflight A pendulum is hanging vertically from the ceiling of an elevator. Initially the elevator is at rest and the period of the pendulum is T. Now the pendulum accelerates upward. The period of the pendulum will now be 1. greater than T 2. equal to T 3. less than T CORRECT “Effective g” is larger when accelerating upward (you feel heavier)

UB, Phy101: Chapter 10, Pg 27 Chapter 10, Preflight Imagine you have been kidnapped by space invaders and are being held prisoner in a room with no windows. All you have is a cheap digital wristwatch and a pair of shoes (including shoelaces of known length). Explain how you might figure out whether this room is on the earth or on the moon

UB, Phy101: Chapter 10, Pg 28 using L=T^2 * g / 4pi^2, solve for gravity. Use the shoelaces to find the T and if g equals 9.8, you are on earth otherwise goodluck All you would have to do is jump up in the air. The gravity on the moon is 1/6 that of earth's. If you were on the moon, you would come down a lot slower.