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 14 cm down from equilibrium and released. After 4 seconds the amplitude has decreased to 13 cm. The spring oscillates 20 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.

Answer 1

Solution:

For damped harmonic motion, the equation is:

D(t) = A₀ · e^(-kt) · cos(ωt)

Finding the values:

1. Initial amplitude: A₀ = 14 cm
2. Angular frequency: ω = 2π(20) = 40π rad/s
3. Decay constant k: Using the fact that amplitude = 13 cm at t = 4:
4. 13 = 14e^(-4k)
5. e^(-4k) = 13/14
6. k = -ln(13/14)/4 ≈ 0.0186

Final equation:

D(t) = 14e^(-0.0186t)cos(40πt) cm

Or exactly: D(t) = 14e^(ln(13/14)·t/4)cos(40πt) cm