Question

Turn object attached to a spring oscillates around the position and is represented by the function Y= 2 cos(2pi(x-0.02)),With time in x seconds. When does a height of 1 foot occur on the interval 0

Answer 1

We are given the function

$y=2\cos (2\pi(x-0.02))$

In order to find the points at which the the height of 1 foot occurs, we consider the equation

$1=2(\text{cos}(2\pi(x-0.02))$

We begin working on it to solve it

$\frac{1}{2}=\cos (2\pi x-0.04\pi)$$\arccos (\frac{1}{2})=2\pi x-0.04\pi$$\frac{\pi}{3}=\pi(2x-0.04)$$\frac{1}{3}=2x-0.04$$x\approx0.19$

In order to find the other point in the interval 0As we can see, the other point that satisfies the equation is x=0.853.

In conclussion, a height of 1 foot occurs at 0.19 and 0.853 seconds.

Another way we can find the second value of x is by noting that

$\cos (\frac{5\pi}{3})=\frac{1}{2}$

And so

$\arccos (\frac{1}{2})=\frac{5\pi}{3}$

With these in mind, we go back to the equation

$\frac{5\pi}{3}=\pi(2x-0.04)$$\frac{5}{3}=2x-0.04$

and so x=0.853