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Oscillations2

Oscillations2 - previous index next Oscillations Michael...

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previous index next Oscillations Michael Fowler 3/24/07 Introduction In this lecture, we will be looking at a wide variety of oscillatory phenomena. After a brief recap of undamped simple harmonic motion, we go on to look at a heavily damped oscillator. We do that before considering the lightly damped oscillator because the mathematics is a little more straightforward—for the heavily damped case, we don’t need to use complex numbers. But they arise very naturally in the lightly damped case, and are great for understanding the driven oscillator and resonance phenomena, as will become apparent in later sections. Brief Review of Undamped Simple Harmonic Motion Our basic model simple harmonic oscillator is a mass m moving back and forth along a line on a smooth horizontal surface, connected to an inline horizontal spring, having spring constant k , the other end of the string being attached to a wall. The spring exerts a restoring force equal to – kx on the mass when it is a distance x from the equilibrium point. By “equilibrium point” we mean the point corresponding to the spring resting at its natural length, and therefore exerting no force on the mass. The in-class realization of this model was an aircar, with a light spring above the track (actually, we used two light springs, going in opposite directions—we found if we just one it tended to sag on to the track when it was slack, but two in opposite directions could be kept taut. The two springs together act like a single spring having spring constant the sum of the two). Newton’s Law gives: 2 2 , or . dx Fm a m k x dt = =− Solving this differential equation gives the position of the mass (the aircar) relative to the rest position as a function of time: 0 () cos ( ) . xt A t ω ϕ = + Here A is the maximum displacement, and is called the amplitude of the motion. 0 t + is called the phase . is called the phase constant : it depends on where in the cycle you start, that is, where is the oscillator at time zero. The velocity and acceleration are given by differentiating x ( t ) once and twice: 00 s in ( ) dx vt A t dt ωϕ = + and

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2 2 2 00 2 () cos ( ) . dxt at A t dt ω ϕ == + We see immediately that this x ( t ) does indeed satisfy Newton’s Law provided 0 is given by 0 /. km = Exercise : Verify that, apart from a possible overall constant, this expression for 0 could have been figured out using dimensions. Energy The spring stores potential energy : if you push one end of the spring from some positive extension x to x + dx (with the other end of the spring fixed, of course) the force – kx opposes the motion, so you must push with force + kx , and therefore do work kxdx. To find the total potential energy stored by the spring when the end is x 0 away from the equilibrium point (natural length) we must find the total work required to stretch the spring from its natural length to an extension x 0 . This means adding up all the little bits of work kxdx needed to get the spring from no extension at all to an extension of x 0 . In other words, we need to do an integral to find the potential energy U ( x 0 ): 0 2 1 2 0 x U x
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Oscillations2 - previous index next Oscillations Michael...

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