1. Warmup A uniform beam 2.20m long with mass m=25.0kg, is mounted by a hinge on a wall as shown. The beam is held horizontally by a wire that makes a 30° angle as shown. The beam supports a mass M = 280kg suspended from its end. Determine the components of the force F H that the hinge exerts and the components tension, F T in the supporting wire. 30° M

2. Harmonic Motion

3. Simple Harmonic Motion When a vibration or oscillation repeats itself, back and forth, over the same path, the motion is periodic Any vibrating system in which the restoring force is directly proportional to the negative of the displacement is said to exhibit simple harmonic motion Such a system is often called a simple harmonic oscillator A mass oscillating on the end of a spring is an example of a simple harmonic oscillator

4. The Motion of an Oscillating Object The motion of an object undergoing simple harmonic oscillation is sinusoidal in nature x(t) = A sin (2πt/T)

5. The Total Energy of a Vibrating System is Constant KE + PE = constant If the maximum amplitude of the motion is x 0 then the energy at any point x is given by: ½ mv 2 + ½ kx 2 = ½ kx 0 2 From this we can solve for velocity: │ v │ = √ [(x 0 2 –x 2 )(k/m)] From Hooke’s law, F = -kx and F =ma, therefore a = -(k/m) x

6. Reference Circle Point P moves with constant velocity v 0 around the circle Point A is the projection of point P on the x axis The motion of A back and forth about point O is second harmonic motion The time for P to go around the circle is T The velocity of point A is v = -v 0 sin θ P A r= x 0 θ θ Displacement x Once around in time T v0v0 v O

7. Reference Circle The period T is: T = 2πr 0 /v 0 = 2πx 0 /v 0 But v 0 is the maximum speed of point A Maximum speed occurs when x = 0 Therefore, v 0 = x 0 √(k/m) Which makes T = 2π√(m/k) And, since f = 1/T, f = (1/2π)√(k/m)

8. Period and Frequency of an Oscillator The period of a simple harmonic oscillator is given by: T = 2π√(m/k) Therefore, since f = 1/T, f = (1/2π)√(k/m)

9. Simple Pendulums For a pendulum the restoring force is F = -mg sin θ, But for small displacements, F = -mg sin θ ~ -mgθ But since x = Lθ we have F ~ (mg/L) x Therefore the motion is essentially harmonic

10. Simple Pendulums But if a pendulum is an harmonic oscillator, then k = mg/L, therefore T = 2π√ (m/mg/L) = 2π√(L/g) and F = 1/T = 1/(2π)√(g/L) The frequency of a pendulum does not depend on the mass of the pendulum bob! Consider the case of large and small children on a swing—the period remains the same

11. In Real Life Most Harmonic Motion is Damped Natural oscillations decrease with time due to friction and other losses Sometimes the damping is so great that the motion does not even appear to be harmonic A = overdamped B = critically damped C = underdamped A shock absorber is a damped oscillator

12. Shock Absorbers Keep a Car From Oscillating

13. Links Java simulationJava simulation—spring Java simulation-oscillator Simple Harmonic Motion and Uniform Circular Motion

14. Do Now (10/8/13): Suppose that a pendulum has a period of 1.5 seconds. How long does it take to make a complete back-and-forth vibration? Is this 1.5 s period pendulum longer or shorter in length than a 1 s period pendulum?

15. Practice: Brainstorm answers to the Conceptual Questions in Chapter 13 (10 min) Complete the Multiple Choice questions in Chapter 13 before the end of class.

16. Do Now (10/9/13): Pass in your HW then wait quietly for instructions