Oscillatory Motion Serway & Jewett (Chapter 15)

M M M Equilibrium position: no net force The spring force is always directed back towards equilibrium (hence called the ‘restoring force’). This leads to an oscillation of the block about the equilibrium position. M For an ideal spring, the force is proportional to displacement. For this particular force behaviour, the oscillation is simple harmonic motion. x F = -kx

SHM: x(t) A = amplitude t f = phase constant w = angular frequency A is the maximum value of x (x ranges from +A to -A). f gives the initial position at t=0: x(0) = A cosf . w is related to the period T and the frequency f = 1/T T (period) is the time for one complete cycle (seconds). Frequency f (cycles per second or hertz, Hz) is the number of complete cycles per unit time.

units: radians/second or s-1  (“omega”) is called the angular frequency of the oscillation.

Velocity and Acceleration a(t) =- w 2 x(t)

Position, Velocity and Acceleration x(t) t v(t) t a(t) t Question: Where in the motion is the velocity largest? Where in the motion is acceleration largest?

Example: SHM can produce very large accelerations if the frequency is high. Engine piston at 4000 rpm, amplitude 5 cm:

Simple Harmonic Motion SHM: We can differentiate x(t): and find that acceleration is proportional to displacement: a(t) = - w 2 x(t) But, how do we know something will obey x=Acos(t) ???

Mass and Spring M Newton’s 2nd Law: F = -kx so x This is a 2nd order differential equation for the function x(t). Recall that for SHM, a = -w 2 x : identical except for the proportionality constant. So the motion of the mass will be SHM: x(t) = A cos (wt + f), and to make the equations for acceleration match, we require that , or (and w = 2p f, etc.). Note: The frequency is independent of amplitude

Example: Elastic bands and a mass Example: Elastic bands and a mass. A mass, m, is attached to two elastic bands. Each has tension T. The system is on a frictionless horizontal surface. Will this behave like a SHO?

Quiz The ball oscillates vertically on a single spring with period T0 . If two identical springs are used, the new period will be longer shorter by a factor of 2 4

Quiz The ball oscillates vertically on a single spring with period T0 . If two identical springs are used, the period will be longer shorter by a factor of 2 4 1

Quiz μs=0.5 B ω=10 s-1 The amplitude of the oscillation gradually increases till block B starts to slip. At what A does this happen? (there is no friction between the large block and the surface) Any A 5 cm 50 cm Not known without mB.

Solution

Look again at the block & spring Energy in SHM M Look again at the block & spring We could also write E = K+U = ½ m(vmax )2

E U, K oscillate back and forth “out of phase” with each other; the total E is constant. n.b.! U, K go through two oscillations while the position x(t) goes through one. U K t T x t v

Suppose you double the amplitude of the motion: 1) What happens to the maximum speed? Doubles 4 x Larger Doesn’t change 2) What happens to the maximum acceleration? Doubles 4 x Larger Doesn’t change 3) What happens to the the total energy? Doubles 4 x Larger Doesn’t change

Summary SHM: (get v, a with calculus) Definitions: amplitude, period, frequency, angular frequency, phase, phase constant. The acceleration is proportional to displacement: a(t) = -w2 x(t) Practice problems, Chapter 15: 3, 5, 11, 19, 23 (6th ed – Chapter 13) 1, 3, 5, 9, 19, 23