Find an equation for the distance, D the end of the spring is below equilibrium in terms of seconds, t.

Question

A spring is attached to the ceiling and pulled 9 cm down from equilibrium and released. The amplitude decreases by 16% each second. The spring oscillates 13 times each second. Find an equation for the distance, D the end of the spring is below equilibrium in terms of seconds, t.

Answer 1

The general form for damped harmonic motion can be written as D(t) = A(t)cos(ωt), where t is time, A(t) is the amplitude function, and ω is the angular frequency.

A(t) = A₀×A_d(t), where the initial amplitude A₀ = 9 cm and A_d(t) represents the decreasing amplitude. Since the amplitude decreases by 16% each second A_d(t) = (1 - 0.16)^t = (0.84)^t, thus the full amplitude function

A(t) = 9(.84)^t.

The period of oscillation T = 2π/ω, or ω = 2π/T. We also know that T = 1/f, where f = 13 Hz is the frequency of oscillation. Putting these equations together we get ω = 2π/(1/13) = 26π.

Finally, the full equation for damped harmonic motion is D(t) = 9(.84)^tcos(26πt)