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Physics 101: Lecture 17 Simple Harmonic Motion. Today’s lecture will cover Textbook Sections 10.1 - 10.4. Simple Harmonic Motion. Period = T (seconds per cycle) Frequency = f = 1/T (cycles per second) Angular frequency = = 2f = 2/T. Simple Harmonic Motion:. x(t) = [A]cos( t)
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Physics 101: Lecture 17Simple Harmonic Motion • Today’s lecture will cover Textbook Sections 10.1 - 10.4
Simple Harmonic Motion Period = T (seconds per cycle) Frequency = f = 1/T (cycles per second) Angular frequency = = 2f = 2/T
Simple Harmonic Motion: x(t) = [A]cos(t) v(t) = -[A]sin(t) a(t) = -[A2]cos(t) x(t) = [A]sin(t) v(t) = [A]cos(t) a(t) = -[A2]sin(t) OR Period = T (seconds per cycle) Frequency = f = 1/T (cycles per second) Angular frequency = = 2f = 2/T For spring: 2 = k/m xmax = A vmax = A amax = A2
relaxed position FX = 0 x x=0 Springs • Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. • FX = -k x Where x is the displacement from the relaxed position and k is the constant of proportionality. (often called “spring constant”)
Springs • Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. • FX = -k x Where x is the displacement from the relaxed position and k is the constant of proportionality. (often called “spring constant”) relaxed position FX = -kx > 0 x • x 0 x=0
Springs • Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. • FX = -k x Where x is the displacement from the relaxed position and k is the constant of proportionality. (often called “spring constant”) relaxed position FX = - kx < 0 x • x > 0 x=0
Springs and Simple Harmonic Motion X=0 X=A; v=0; a=-amax X=0; v=-vmax; a=0 X=-A; v=0; a=amax X=0; v=vmax; a=0 X=A; v=0; a=-amax X=-A X=A
x +A CORRECT t -A 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
x +A CORRECT t -A 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
Review: Period of a Spring For simple harmonic oscillator = 2f = 2/T For mass M on spring with spring constant k
x = R cos q = R cos (wt) since q = wt x 1 1 R 2 2 3 3 0 y 4 6 -R 4 6 5 5 What does moving in a circlehave to do with moving back & forth in a straight line ?? x 8 8 q R 7 7
m x x=0 Potential Energy of a Spring Where x is measured fromthe equilibrium position PES Analogy: marble in a bowl: PE KE PE KE PE “height” of bowl x2 x 0
m Same thing for a vertical spring: y y=0 Where y is measured fromthe equilibrium position PES y 0
CORRECT 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 12. Case 23. Same
Chapter 10, Preflight PE = 0 KE = KEMAX PE = 1/2kx2KE = 0 same for both samefor both x=+A x=0 x=-A x=+A x=0 x=-A
Pendulum Summary DEMO • For “small oscillation”, period does not depend on • mass • amplitude
CORRECT 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 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 “Effective g” is larger when accelerating upward (you feel heavier)
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
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.