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

There are two components to the movement, oscillation and dampening.

For oscillation, the spring position will follow a sine or cosine function, since it has maximum value at t = 0, we'll use the cosine function. Given an oscillation function, O(t) = O₀cos(wt), the period of the oscillation is 2pi / w. Since the spring oscillates 16 times per second, the period of each oscillation is 1/16 sec. So we have 2pi / w = 1/16 sec, or w = 32pi /sec. The initial value is 8cm, so O(t) = 8 cos(32pi t), O(t) in cm, t in seconds.

For dampening, we have the amplitude declining by 7% each second, or A_t+1 = A_t x (0.93), repeating each second. thus starting with the unit value, A₀ = 1, we have after t seconds: A_t = 1 x (0.93)^t

We multiply the oscillation equation by the amplitude equation to get:

D(t) = 8 x (0.93)^t x cos(32pi t), D in cm, t in seconds.