Chapter 11: Harmonic Motion

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Presentation transcript:

Chapter 11: Harmonic Motion Elastic Potential Energy energy stored in a stretched/compressed spring Force: Hooke’s Law F = kx F = 0 F = k x/2 F = k x x unstretched half stretched stretched

A horizontal spring had a force constant of 90 N/m. A mass of 1 A horizontal spring had a force constant of 90 N/m. A mass of 1.5 kg is attached free end. The spring is then compressed by 50 cm, and then released. What is the speed of the mass when it returns to the equilibrium position?

Simple Harmonic Motion: oscillations Period T = time for one complete oscillation frequency f = number of oscillations per time usually number per second (1 cycle/sec = 1 Hertz = 1Hz) f = 1/T for a restoring force (equilibrium) + inertia Fr = kx plus F = ma

Position, speed and acceleration in simple harmonic motion x maximum speed at x = 0 Amplitude A = xmax v = 0 when x = xmax x = A cos 2ft = A cos t t

Example: An object undergoes SHM with a frequency of 20 Hz and a maximum speed of 2.5 m/s. What is the amplitude of the motion? What is the object’s displacement when its speed is 1.5 m/s?

mass m on a string of length L The Simple Pendulum mass m on a string of length L x s mg T Fnet L Example: How long should a pendulum be in order to have a period of 1.0 s?