Find an equation for each world problem.

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.

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.

A population of rabbits oscillates 33 above and below average during the year, hitting the lowest value in January (t = 0). The average population starts at 700 rabbits and increases by 170 each year. Find an equation for the population, P, in terms of the months since January, t.

Answer 1

Question 1: Spring pulled 14 cm, amplitude decreases to 13 cm after 4 sec.

Step 1: General form.

1. For a damped spring, the equation would be D(t) = A₀e^-kt cos (ωt) → We use cosine due to it's released from macimum displacement.

Step 2: Use the given values.

1. The initial amplitude is A₀ = 14
2. And after 4 sec, the amplitude is 13. So that means it would be: 14e^-4k = 13
3. Now solve for k, which will be: e^-4k = 13/14 → k = -1/4 ln (13/14)
4. The oscillations would be 20 per second: ω = 2π(20) = 40π

That means the final equation would be: D(t) = 14e^{-(-1/4 ln (13/14)) t} cos (40πt)

Question 2: Spring pulled 9 cm, amplitude decreases by 16% per second

Step 1: Expoenential decay problem

1. That means that if amplitude decreases by 16% each second, it would be: A(t) = 9(0.84)^t

Step 2: Use the Aungular Frequency

1. The oscillates 13 times per second, which means it would be: ω = 2π(13) = 26π

That means that the final equation would be D(t) = 9(0.84)^t cos (26πt)

Question 3: Rabbit population oscillates yearly with increasing average.

Step 1: Average (finding the midline)

1. It starts at 700 rabbits, and increases by 170 per year. That means that it is 170/12 per month. So that means that the midline would be: M(t) = 700 + 170/12t

Step 2: Amplitude

1. Oscillates 33 above and below average.

Step 3: Time period

1. 1 year = 12 months,
2. That means the equation would be: ω = 2π/12 = π/6

Step 4: Phase (finding the lowest value at t = 0

1. The lowest value would be using negative cosine: -33 cos (π/6 t)

So that means that the final equation would be: P(t) = 700 + 170/12 t - 33 cos (π/6 t)