Physics 101: Lecture 22 Simple Harmonic Motion Today’s lecture will cover Textbook Sections 10.4 - 10.6 1

Review: Ideal 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

Review: Simple Harmonic Motion Uniform circular motion <-> Motion of an object attached to an ideal spring Period = T (seconds per cycle) Frequency = f = 1/T (cycles per second) Angular frequency =  = 2f = 2/T

Simple Harmonic Motion: Quick Review 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) OR xmax = A vmax = A amax = A2 Period = T (seconds per cycle) Frequency = f = 1/T (cycles per second) Angular frequency =  = 2f = 2/T

Review: Period of a Spring For simple harmonic oscillator  = 2f = 2/T For mass M on spring with spring constant k Demos: A,m,k dependence

Concept Question x +2A t -2A If the amplitude of the oscillation (same block and same spring) was doubled, how would the period of the oscillation change? (The period is the time it takes to make one complete oscillation) 1. The period of the oscillation would double. 2. The period of the oscillation would be halved 3. The period of the oscillation would stay the same CORRECT x +2A t -2A

Potential Energy of a Spring m x x=0 Where x is measured from the equilibrium position PES x

Same thing for a vertical spring: y y=0 m Where y is measured from the equilibrium position PES y

Etotal = 1/2 Mv2 + 1/2 kx2 = constant In either case... m y y=0 m x x=0 Etotal = 1/2 Mv2 + 1/2 kx2 = constant KE PE KEmax = 1/2Mv2max=1/2M2A2 =1/2kA2 PEmax = 1/2kA2 Etotal = 1/2kA2

Concept Question 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

Concept Question PE = 0 KE = KEMAX PE = 1/2kx2 KE = 0 same for both x=-A x=0 x=+A x=-A x=0 x=+A

Concept Question The same would be true for vertical springs... PE = 1/2k y2 Y=0 PE = 1/2k y2 Y=0

Pendulum For “small oscillation”, period does not depend on mass amplitude Demos: M,A,L dependence

Concept Question 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

“Effective g” is larger when accelerating upward Concept Question 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)