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Physics 101: Lecture 22 Simple Harmonic Motion

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

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Physics 101: Lecture 22 Simple Harmonic Motion

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

  2. 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 xWhere xis 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

  3. 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

  4. 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 Period = T (seconds per cycle) Frequency = f = 1/T (cycles per second) Angular frequency =  = 2f = 2/T xmax = A vmax = A amax = A2

  5. 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

  6. CORRECT Concept Question 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 halved3. The period of the oscillation would stay the same x +2A t -2A

  7. m x x=0 Potential Energy of a Spring Where x is measured fromthe equilibrium position PES x 0

  8. m Same thing for a vertical spring: y y=0 Where y is measured fromthe equilibrium position PES y 0

  9. CORRECT 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 12. Case 23. Same

  10. Concept Question 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

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

  12. CORRECT 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

  13. CORRECT 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 “Effective g” is larger when accelerating upward (you feel heavier)

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