Find an equations for the populations.

Question

A population of rabbits oscillates 24 above and below average during the year, hitting the lowest value in January (t = 0). The average population starts at 800 rabbits and increases by 7% each month. Find an equation for the population, P, in terms of the months since January, t.

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

Question 1: Rabbit Population (oscillation with growing average)

It is given that:

1. Oscillates 24 above and below average → amplitude A = 24
2. Lowest value in January t = 0 → use negative cosine
3. Average starts at 800
4. Average increases by 7% each month
5. And the time period is: 1 year = 12 months

Step 1: Average (finding the midline)

1. Using the exponential growth equation, it would be: M(t) = 800 (1.07)^t

Step 2: Angular frequency

1. ω = 2π/12 = π/6

Step 3: Oscillation termα

1. Loest at t = 0 → -24 cos (π/6 t)

That means that the final equation is P(t) = 800 (1.07)^t - 24 cos (π/6 t)

Question 2: Damped spring motion

Based on the given information:

1. Initial amplitude is 14 cm.
2. After 4 sec, the amplitude it 13 cm
3. Oscillates 20 times per second
4. Amplitude decays exponenetially
5. And it is released from macimum displacement, so that means it is cosine

Step 1: Using the general mode.

1. D(t) = A₀​e^-kt cos (ωt)

Step 2: Fionding the decay constant k

1. 14e^-4k = 13
2. e^-4k = 13/14
3. k = -1/4 ln (13/14)

Step 3: Angular frequency

1. ω = 2π(20) = 40π

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