MATH 282 - DE οͺ 4.2 Sinusoidal Forcing Today we are going to study forced equations where the forcing function is sinusoidal (either sine or cosine). Let’s apply the guessing technique to sinusoidally forced linear equations: Example. Let’s calculate the general solution to the equation: π2 π¦ ππ¦ + + 2π¦ = cos 2π‘ ππ‘ 2 ππ‘ Solution: 1 Here are the graphs of three solutions: Here is the graph of the steady-state solution: A little translation: Consider the 2nd order linear forced equation π2π¦ ππ¦ π 2 +π + π π¦ = π(π‘) ππ‘ ππ‘ where π and π are positive and π ≥ 0. Engineering Terminology: ο· forced response – any solution to the forced equation. ο· steady-state response – behavior of the forced response over the long term. ο· natural (or free) response – any solution of the associated homogeneous equation. Why are initial conditions essentially irrelevant? When we guess a solution of the form π¦π (π‘) = πΌπ πππ‘ and compute the complex number πΌ, we have essentially determined everything we need to know about the steady-state solution. Euler’s formula gives us a nice way of determining the amplitude, and phase-angle of the steady-state solution immediately from πΌ. 2 Why we prefer to calculate the steady-state solution using complex numbers? Recall that we calculated the steady-state solution 1 π¦π (π‘) = − (cos 2π‘ − sin 2π‘) 4 for the equation π 2 π¦ ππ¦ + + 2π¦ = cos 2π‘, ππ‘ 2 ππ‘ and we did so using computations that involved complex numbers. In fact, we found π¦π (π‘) as the real part of 1 π¦π (π‘) = − (1 + π)π (2π)π‘ . 4 The complex number 1 πΌ = − (1 + π) 4 Tells us everything we need to know about the steady-state solution. In order to see why, we use polar coordinates in the complex plane ( see pp. 745-747 in Appendix C of the text). 3