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The equation that models the height above or below equilibrium in inches, h, of a spring over time in seconds, t, is h equals negative 15 cosine left parenthesis Start Fraction 2 pi over 5 End Fraction t right parenthesis. At which times will the spring be at a height of 8 in. above equilibrium? Select 2 that apply.
(1 point)
Responses

4.2 seconds

1.7 seconds

2.9 seconds

0.8 seconds

3.3 seconds

Answer 1

To find the times when the spring's height \( h \) is 8 inches above equilibrium, we need to set the equation equal to 8 inches:

\[
-15 \cos\left(\frac{2\pi}{5} t\right) = 8
\]

First, we can solve for \( \cos\left(\frac{2\pi}{5} t\right) \):

1. Divide both sides by -15:
\[
\cos\left(\frac{2\pi}{5} t\right) = -\frac{8}{15}
\]

Next, we will take the inverse cosine of both sides:
\[
\frac{2\pi}{5} t = \cos^{-1}\left(-\frac{8}{15}\right)
\]

Calculating \( \cos^{-1}\left(-\frac{8}{15}\right) \):

Using a calculator, you will find:
\[
\cos^{-1}\left(-\frac{8}{15}\right) \approx 2.5373 \text{ radians}
\]

Since cosine is periodic, we also need to consider that cosine has the same value at \( 2\pi - \theta \):
\[
\frac{2\pi}{5} t = 2\pi - \cos^{-1}\left(-\frac{8}{15}\right) \approx 2\pi - 2.5373 \approx 3.7453 \text{ radians}
\]

Now, we can solve for \( t \):

1. For the first solution:
\[
\frac{2\pi}{5} t = 2.5373 \\
t = \frac{5}{2\pi} \cdot 2.5373 \approx \frac{5 \cdot 2.5373}{6.2832} \approx 2.022
\]

2. For the second solution:
\[
\frac{2\pi}{5} t = 3.7453 \\
t = \frac{5}{2\pi} \cdot 3.7453 \approx \frac{5 \cdot 3.7453}{6.2832} \approx 2.978
\]

Now we need to check the provided options to see which are closest:

- 4.2 seconds
- 1.7 seconds
- 2.9 seconds (close to 2.978)
- 0.8 seconds
- 3.3 seconds

Based on our calculations, the answers should be:

- **2.9 seconds**
- There would be one more solution due to the periodicity of the cosine function, so we also need to account for additional cycles.

Since \( \cos(x) = -\frac{8}{15} \) gives us a periodic solution, we can calculate another solution by adding the period:
\[
3.7 + T = 3.7 + 5 \cdot 1 = 8.7 \text{ or similar values}
\]

Yet they are likely to match with **2.9 seconds** and any nearby value through period-based addition.

Thus, the only option available specifically from given segments is **2.9 seconds**, with additional answers dependent upon clear intervals being tested elsewise.

For the exercise context, you can still check against further resources if seeking deeper cycle outcomes.