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).

 

[Architeuthis Dux]

I would approach this problem by first defining the variables and parameters involved. Let's say the distance from the ground to the tip of the spring is represented by h, and the distance from the ceiling to the tip of the spring is represented by d. We also need to define the time variable, which we can call t.

Next, we can use the equation for simple harmonic motion to model the distance from the ground in terms of time:

h = d + A cos(ωt + φ)

Where A is the amplitude, ω is the angular frequency, and φ is the phase constant.

In this case, the amplitude decreases by 5% every second, so we can represent it as A = A0(0.95)^t, where A0 is the initial amplitude when the spring is released. We also know that the period is pi, so we can use the relationship T = 2π/ω to solve for ω, which gives us ω = 2π/pi.

Putting all of this together, we get the equation:

h = d + A0(0.95)^t cos(2πt/pi + φ)

To solve for the phase constant, we can use the initial conditions. When the spring is released, its initial position is 15 meters from the ground, so we can set h = 15 and t = 0 in the equation above, which gives us:

15 = d + A0 cos(φ)

We also know that the initial position of the spring is d, so we can set h = d in the equation above, which gives us:

d = d + A0 cos(φ)

Solving for cos(φ), we get:

cos(φ) = (d - 15)/A0

Substituting this into our original equation, we get the final equation:

h = d + A0(0.95)^t cos(2πt/pi + cos^-1((d-15)/A0))

This equation models the distance from the ground in terms of time for the given spring problem. It takes into account the decreasing amplitude and the initial conditions.