How Do You Model the Motion of a Spring Attached to a Ceiling with Energy Loss?

[Hockeystar]

Homework Statement

A spring is attached to a ceiling with a distance of 20 meters from the ground to the tip of the spring. The spring is brought down 5 meters down (This means 15 meters from the ground) and then released. The spring loses 5% of its energy each second. Assume a period of pi. Determine an equation to model the distance from ground in terms of time.

Homework Equations

Trig function that exponentially gets smaller amplitude. I'm guessing the period would get shorter.

The Attempt at a Solution

My guess is f(x) = .95^(x)(-5cos(pi(1.05^x))) + 16

Can someone verify/correct me.

 

[HallsofIvy]

You are told that the period is pi, a constant! So, immediately, your "guess" that the period would get shorter is incorrect. The decrease in energy is a decrease in the amplitude of the motion, not the period. A function of period pi is either sin(2t) or cos(2t). Further the motion will be symmetric around the "rest position" which you are told is 20 meters from the ground. Letting x be the distance from the ground to the tip of the spring, x(t)= 20- A(t)cos(2t). The spring will be at its lowest point when t is a multiple of pi: t= n pi, where x(2n pi)= 20- A(n pi). The reason I am looking at the lowerst point is that the speed, and so kinetic energy, will be 0 there so the total energy is just the potential energy which, relative to the ground, is mgx= 20mg- mgA(n pi). In n pi seconds, it will have lost .05^{n pi} of its original energy which was 15mg: Solve mg(20- A(n pi))= .05^n\pi(15 mg) or, more generally, 20- A(t)= .05^{n pi}(15), for A(t).