A spring is attached to the ceiling and pulled 20cm down from

Question

A spring is attached to the ceiling and pulled 20 cm down from equilibrium and released. The amplitude decreases by 13% each second. The spring oscillates 7 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

An oscillating spring indicates a periodic equation. Initially, let's ignore the amplitude decrease.

Since at time zero, the spring is below the equilibrium point at the maximum distance, the easiest function to use is a negative cosine function:

y = -Acos(Bt) + D (there is no phase shift so C = 0)

The amplitude is 20 cm so we can use height "y" in cm and plug in A = 20:

y = -20cos(Bt) + D

The "B" term is calculated from the period. B = 2π/Period and the period = 1/7 second so we can let time "t" be in seconds and use B = 2π/(1/7) = 14π so:

y = -20cos(14πt) + D

Since we are looking for an equation for distance (D) that is below the equilibrium point, the "D" in our equation (different "D") is zero, But the wording in your problem is difficult to understand. Going SRICTLY by the words "distance the end of the spring is below equilibrium" then the t = 0 value should be positive meaning the equation should be y = 20cos(14πt) but this seems confusing because then distances above the equilibrium point would be negative. So, I'll leave the negative sign in and define the equation:

D(t) = -20cos(14πt)

Where time "t" is in seconds and the distance "D" is is in cm and is the distance from the equilibrium point where negative means below the equilibrium point.

Now, put in the damping of 13%. We can use the exponential function A = A₀e^kt and we can calculate "k" by using A = 87, A₀ = 100, and t = 1:

87 = 100e^k(1)

0.87 = e^k

ln(0.87) = ln(e^k)

-0.139262 = k

Now we can multiply our periodic function by the damping function:

D(t) = -20e^-0.139262tcos(14πt)