Question

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 all that apply.
(1 point)
Responses

4.2 seconds

1.7 seconds


2.9 seconds
2.9 seconds

0.8 seconds

3.3 seconds

Answer 1

To find the times when the spring is at a height of 8 inches above equilibrium, we start with the given equation:

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

To find the times when \( h(t) = 8 \), we set the equation equal to 8:

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

Dividing both sides by -15 gives:

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

Next, we need to find the angles where the cosine equals \(-\frac{8}{15}\). The cosine function is negative in the second and third quadrants. Therefore, we can write:

\[
\frac{2\pi}{5} t = \cos^{-1}\left(-\frac{8}{15}\right) + 2k\pi \quad \text{and} \quad \frac{2\pi}{5} t = -\cos^{-1}\left(-\frac{8}{15}\right) + 2k\pi
\]

for \( k = 0, \pm1, \pm2, \ldots \)

Let's denote \( \theta = \cos^{-1}\left(-\frac{8}{15}\right) \).

Now we can calculate \( \theta \):

Using a calculator,
\[
\theta \approx 2.46 \text{ (in radians)}
\]

Now we substitute back to find \( t \):

1. For the second quadrant:
\[
\frac{2\pi}{5} t = \theta \implies t = \frac{5\theta}{2\pi} \approx \frac{5 \times 2.46}{2\pi} \approx 1.96 \text{ seconds}
\]

2. For the third quadrant:
\[
\frac{2\pi}{5} t = 2\pi - \theta \implies t = \frac{5(2\pi - \theta)}{2\pi} \approx \frac{5(2\pi - 2.46)}{2\pi} \approx 3.33 \text{ seconds}
\]

These calculations yield approximate times of 1.96 seconds and 3.33 seconds.

Looking at the responses provided:

- **1.7 seconds**
- **2.9 seconds**
- **0.8 seconds**
- **3.3 seconds**

The only matching time is:

**3.3 seconds**

So, the correct answer is **3.3 seconds.**