Trigonometry question

Question

Suppose an object attached to a coiled spring is pulled down a distance of 18 in from its rest position and then released. If the time for one oscillation is 11 sec, develop a model that relates the displacement d(t) of the object from its rest position after time t (in seconds). Assume no friction.

d(t)=

Answer 1

We know that the amplitude is 18 and that at t = 0 that is its value.

So, Y = 18 cos ωt is a good start with the greek letter omega multiplying t.

But omega = 2pi/T, where T = the period, which is given as 11.

So, substituting for omega we get y = ampllitude = 18 cos (2Pit/11)

Answer 2

The amplitude of the equation will equal 18 in. period equals 11 sec. Since we are pulling the spring down, we want to use cos and our function will be negative.

In general d(t) = -Acos(Bt)

We know A = 18.

B equals 2pi/period. So we get 2pi/11 is our B value.

That gives us d(t) = -18cos(2pi/11)t