To solve for the times when the spring is at a height of [tex]\( h = 8 \)[/tex] inches above equilibrium, we start with the given equation:
[tex]\[ h = -15 \cos \left(\frac{2 \pi}{5} t\right) \][/tex]
We want to find the values of [tex]\( t \)[/tex] when [tex]\( h = 8 \)[/tex]:
[tex]\[ 8 = -15 \cos \left(\frac{2 \pi}{5} t\right) \][/tex]
First, we isolate the cosine term by dividing both sides by [tex]\(-15\)[/tex]:
[tex]\[ \frac{8}{-15} = \cos \left(\frac{2 \pi}{5} t\right) \][/tex]
This simplifies to:
[tex]\[ -\frac{8}{15} = \cos \left(\frac{2 \pi}{5} t\right) \][/tex]
Now, we need to find the angle whose cosine is [tex]\(-\frac{8}{15}\)[/tex]. The principal angle [tex]\( \theta \)[/tex] for which the cosine is a certain value can be found using the inverse cosine function:
[tex]\[ \theta = \cos^{-1} \left(-\frac{8}{15}\right) \][/tex]
Once we have [tex]\( \theta \)[/tex], we need to solve for [tex]\( t \)[/tex] in the original equation argument:
[tex]\[ \frac{2 \pi t}{5} \][/tex]
We have two solutions:
1. [tex]\( \frac{2 \pi t}{5} = \theta \)[/tex]
2. [tex]\( \frac{2 \pi t}{5} = 2\pi - \theta \)[/tex]
Solving for [tex]\( t \)[/tex] in each case:
1. [tex]\( t_1 = \frac{5 \theta}{2 \pi} \)[/tex]
2. [tex]\( t_2 = \frac{5 (2\pi - \theta)}{2 \pi} = 5 - \frac{5 \theta}{2 \pi} \)[/tex]
Next, we compare these [tex]\( t \)[/tex]-values with the given times [tex]\( 4.2 \)[/tex] seconds, [tex]\( 1.7 \)[/tex] seconds, [tex]\( 2.9 \)[/tex] seconds, [tex]\( 0.8 \)[/tex] seconds, and [tex]\( 3.3 \)[/tex] seconds to see which of them satisfy our equations.
Upon evaluating:
- [tex]\( t_1 \approx 1.7 \)[/tex] seconds
- [tex]\( t_2 \approx 3.3 \)[/tex] seconds
Therefore, the times when the spring is at a height of [tex]\( 8 \)[/tex] inches are:
1.7 seconds and 3.3 seconds.
Thus, the correct answers are:
[tex]\[
\boxed{1.7 \ \text{seconds}, \ 3.3 \ \text{seconds}}
\][/tex]
These are the times at which the spring will be at a height of 8 inches above equilibrium.
