A spring is attached to the ceiling and pulled 8 cm down from equilibrium and released.

Question

A spring is attached to the ceiling and pulled 8 cm down from equilibrium and released. The amplitude decreases by 7% each second. The spring oscillates 16 times each second. Find an equation for the distance, D the end of the spring is below equilibrium in terms of seconds, t.

D(t) =

Answer 1

The graph is oscillating but it's outlining an exponential function

The general formula for this would be:

D(t) = ab^tcos(ωt).

(The spring starts out stretched and then released so we want a cosine function with no phase shift)

The spring is being pulled down -8 so at t = 0, D(t) = -8. Plug this in:

-8 = ab⁰cos( ω(0) ) = a

a = -8

So D(t) = -8b^t cos(ωt)

Next, it's decreasing by 7% every second so b is equal to 1 - 7% = 0.93.

Lastly, you can get ω to be 2π (16) because it oscillates 16 times every second