 # A mass hanging from a vertical spring is in simple harmonic Cheexorgeny 2021-12-14 Answered
A mass hanging from a vertical spring is in simple harmonic motion as given by the following position function, where t is measured in seconds and s is in inches: $s\left(t\right)=6\mathrm{cos}\left(\pi t+\frac{\pi }{4}\right)$.
Round answers to 2 decimal places.
a. Determine the position of the spring at $t=2.5s$
b. Find the velocity of t spring $t=2.5s$
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Step 1
The given position function is:
$s\left(t\right)=-6\mathrm{cos}\left(\pi t+\frac{\pi }{4}\right)$
where t is in seconds and $s\left(t\right)$ is in inches.
(a)
To determine $s\left(2.5\right)$.
That is position at $t=2.5$
So,
$s\left(2.5\right)=-6\mathrm{cos}\left(\pi \left(2.5\right)+\frac{\pi }{4}\right)$
$=-6\mathrm{cos}\left(\frac{5\pi }{2}+\frac{\pi }{4}\right)$
$=-6\mathrm{cos}\left(\frac{11\pi }{4}\right)$
$\approx 4.24$

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Step 2
b)
Now velocity is given by:
$v\left(t\right)={s}^{\prime }\left(t\right)$
$=\frac{d}{dt}\left(s\left(t\right)\right)$
$=\frac{d}{dt}\left(-6\mathrm{cos}\left(\pi t+\frac{\pi }{4}\right)\right)$
$=-6\frac{d}{dt}\left(\mathrm{cos}\left(\pi t+\frac{\pi }{4}\right)\right)$
$=-6\left(-\mathrm{sin}\left(\pi t+\frac{\pi }{4}\right)×\pi \right)$
$=6\pi \mathrm{sin}\left(\pi t+\frac{\pi }{4}\right)$
So,
$v\left(2.5\right)=6\pi \mathrm{sin}\left(\frac{5\pi }{2}+\frac{\pi }{4}\right)$
$=6\pi \mathrm{sin}\left(\frac{11\pi }{4}\right)$
$\approx 13.33$