Consider the equation d 2 dt2 + k m y = 0 for the motion of a simple harmonic oscillator. (a) Consider the function y(t) = cos βt. Under what conditions on β is y(t) a solution to the differential equation? (b) What initial condition (t = 0) in the yv-plane corresponds to this solution? (c) In terms of k and m, what is the period of this solution? (d) Sketch the solution curve in the yv-plane that corresponds to this solution.

d? maybe

step-by-step explanation:

step-by-step explanation:

is their an equation anywhere i can take a   look at it and see if i am right.

18 units²

step-by-step explanation:

first let's find altitude bd using the distance formula

d = √((y₂ - y₁)² + (x₂ - x₁)²)

bd = √((0 - -3)² + (0 - 3)²)

bd = √18

given the placement of points a and c, we can denote that point d bisects segment ac.

we will call ac the base.

using the distance formula

ac = √((0 - 6)² + (-6 - 0)²)

ac = √72

knowing our base and our height, we can plug these into our triangle area formula

area = (b * h)/2

area = (√18 * √72) / 2

area = (√1296) / 2

area = 36 / 2

area = 18 units²