Please answer part B only. Show that the general solution of d 2x /dt 2 + 2 lamb

Question

Please answer part B only.

Show that the general solution of d 2x /dt 2 + 2 lambda dx/dt + omega 2 x = f 0 sin gamma t is x(t) = Ae -lambda t sin ( + phi) + F 0/ sin (gamma t + theta), where A = and the phase angles phi and theta are, respectively, defined by sin phi = c 1/A, cos phi = c 2/A and sin theta = -2 lambda gamma / cos theta = omega 2 - gamma 2/ The solution in part (a) has the form x(t) = x C(t) + x P(t). Inspection shows that x C(t) is transient, and hence for large values of time, the solution is approximated by x P(t) = g(gamma) sin(gamma t + theta), where g(gamma) = F 0/ Although the amplitude g(gamma) of x p(t) is bounded as t rightarrow infinity, show that the maximum oscillations will occur at the value gamma 1 = . What is the maximum value of g? The number 2 pi is said to be the resonance frequency of the system.

Explanation / Answer

Please excuse me for using w= and y=. It's much easier for me to type the solution that way!

We wish to maximize g(y). Notice that the numerator of g(y) is constant, so we may maximize g by minimizing the denominator. The denominator is always positive (any real number squared is greater than zero), so this is the same as minimizing the square of the denominator.

Define d(y)=(w2-y2)2+42y2. We wish to minimize d. We start my differentiating with respect to y:

d'(y)= 2(w2-y2)(-2y)+82y=0. Note, we set the derivative equal to zero, as we would for any critical point problem.

Then, we divide through by y, assuming y is not equal to zero, and simiplify:

-4(w2-y2)+82=0

22=w2-y2.

y2=w2-22

So, y=(w2-22)

Now, to find the maximal value of g, we substitute back in (w2-22) for y2:

d(y)=(w2-(w2-22))2+42(w2-22)

d(y)=44+42w2-84 = 42(w2-2)

And plugging back ingo g(y) give

gmax=F0/[2(w2-2)]

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