A spring is attached to the ceiling and pulled 16 cm down from equilibrium and released

Question

A spring is attached to the ceiling and pulled 16 cm down from equilibrium and released. After 4 seconds the amplitude has decreased to 7 cm. The spring oscillates 6 times each second. Assume that the amplitude is decreasing exponentially. Find an equation for the distance, D the end of the spring is below equilibrium in terms of seconds, t.

D(t) =

please explain clear its due in 1 hour hw

Answer 1

Were going to model this as simple harmonic motion, so the displacement equation would be:

D(t) = A(t) * sin(2*π*f *t)

A is the amplitude and f is the frequeny of oscillation.

They tell us in the problem that the amplitude A decreases exponentially, so we can model A(t) as:

A(t) = A_o*e^-kt , where A_o is the amplitude at t=0

First, we will solve for k in this equation:

The problem says that it starts at 16cm, and then at 4 seconds it goes to 7cm

So we can say 7cm = 16cm * (e^-k(4seconds))

7/16 = e^-4k

ln(7/16) = -4k

So k = ln(7/16)/-4 = 0.21, with sig figs it's 0.2

So we now have A(t) =A_o*e^-02t

We have the frequency given to us at 6 s^-1 so we can plug that in for f

So our total equation is D(t) = 16e^-0.2tsin(12πt)