Wile E. Coyote has devised a sure bet means of catching the road runner that inv

Question

Wile E. Coyote has devised a sure bet means of catching the road runner that involves a giant spring purchased from ACME products, a safe and a box of bird seed. When the roadrunner stops to eat the free bird seed, the safe will be released killing the flightless aves once and for all. When the spring then retracts carrying the safe with it, the coyote can amble down the cliff and remove the flattened carcass for his barbecue dinner.

(a) According to factory specs, a 250 Newton force will displace the spring 6 meters. The safe has a mass of 1500 kilograms, and is released 20 meters above the equilibrium point. If the roadrunner is located precisely at the distance of maximal extension of the spring-safe system (i.e. the minimum position of the solution function y=x(t) ), how long will the coyote have to recover the dead roadrunner before the safe slams back down onto his head on its second cycle? (You must write and solve an appropriate differential equation to receive credit for this problem.)

Explanation / Answer

The elastic force is involved here. Let us assume the Hooke law F = k x; the elastic constant k = F / x = 250 N / 6 m = 41.6 N m^-1

The differential equation of motion is F = m a <=> -k x = m d^2 x/ dt^2. It tells you that the solution x(t) is such that its second derivative is proportional to the solution itself. Among the elementary functions, the exponential does behave in such a way since its derivatives are still exponential as well. So, let us try with x = Ae^(p t). Its second derivative is d^2 x / dt^2 = p^2 A e^(pt). Thus, substituting them into the equation, one is left with - k = m p^2, so that p^2 = - k/m => p= i sqrt(k/m) = i omega, where we have defined omega = sqrt(k/m), and i = sqrt(-1). The solution does exist, and it is x(t) = A e^(i omega t) = a e^i(omega t + phi) since, in general, A = a e^i phi is a complex number. The oscillation period is T =2pi/omega= 2pi sqrt(m / k) = 37 s. Let us take the real aprt of the solution x(t) = a cos omega t, so that at t=0, x0 = a. In our case, x0= 20 m.Thus, one has to impose 6 m = 20 m cos omega t* => t* = (1/omega) arccos(6/20) = (1/omega) 1.26 rad = 7.6 s.

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