A spring is oscillating vertically according to the equation y = 3cos(2t-pi), where y is the displacement in feet and t is time in seconds. At what time will thespring be 2 ft. high? Round your answer to the nearest tenth.

A spring is oscillating vertically according to the equation y 3cos2tpi where y is the displacement in feet and t is time in seconds At what time will thespring class=

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Answer: 25.7 seconds

Explanation

Equation

[tex]y=3\cos(2t-\pi)[/tex]

where y is the displacement in feet and t is time in seconds.

We are asked to find t when y = 2ft, meaning that we have to solve the equation for t:

0. Replacing the information given by setting the equation to 2:

[tex]3\cos(2t-\pi)=2[/tex]

2. Dividing both sides of the equation against 3 and simplifying:

[tex]\frac{3\cos(2t-\pi)}{3}=\frac{2}{3}[/tex][tex]\cos(2t-\pi)=\frac{2}{3}[/tex]

2. Applying secant (cos⁻¹) to both sides:

[tex]\cos^{-1}(\cos(2t-\pi)=\cos^{-1}(\frac{2}{3})[/tex][tex](2t-\pi)=\cos^{-1}(\frac{2}{3})[/tex]

3. Adding π to both sides of the equation:

[tex]2t-\pi+\pi=\cos^{-1}(\frac{2}{3})+\pi[/tex][tex]2t=\cos^{-1}(\frac{2}{3})+\pi[/tex]

4. Dividing both sides of the equation against two:

[tex]\frac{2t}{2}=\frac{\cos^{-1}(\frac{2}{3})+\pi}{2}[/tex][tex]t=\frac{\cos^{-1}(\frac{2}{3})+\pi}{2}[/tex]

5. Simplifying we get:

[tex]t\approx\frac{51.33}{2}=25.67[/tex]