# Difference between revisions of "009A Sample Midterm 1, Problem 5"

The displacement from equilibrium of an object in harmonic motion on the end of a spring is:

$y={\frac {1}{3}}\cos(12t)-{\frac {1}{4}}\sin(12t)$ where  $y$ is measured in feet and  $t$ is the time in seconds.

Determine the position and velocity of the object when  $t={\frac {\pi }{8}}.$ Foundations:
What is the relationship between the position  $s(t)$ and the velocity  $v(t)$ of an object?
$v(t)=s'(t)$ Solution:

Step 1:
To find the position of the object at  $t={\frac {\pi }{8}},$ we need to plug  $t={\frac {\pi }{8}}$ into the equation  $y.$ Thus, we have
${\begin{array}{rcl}\displaystyle {y{\bigg (}{\frac {\pi }{8}}{\bigg )}}&=&\displaystyle {{\frac {1}{3}}\cos {\bigg (}{\frac {12\pi }{8}}{\bigg )}-{\frac {1}{4}}\sin {\bigg (}{\frac {12\pi }{8}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {1}{3}}\cos {\bigg (}{\frac {3\pi }{2}}{\bigg )}-{\frac {1}{4}}\sin {\bigg (}{\frac {3\pi }{2}}{\bigg )}}\\&&\\&=&\displaystyle {0-{\frac {1}{4}}(-1)}\\&&\\&=&\displaystyle {{\frac {1}{4}}{\text{ ft}}.}\end{array}}$ Step 2:
Now, to find the velocity function, we need to take the derivative of the position function.
Thus, we have
${\begin{array}{rcl}\displaystyle {v(t)}&=&\displaystyle {y'}\\&&\\&=&\displaystyle {{\frac {-1}{3}}\sin(12t)(12)-{\frac {1}{4}}\cos(12t)(12)}\\&&\\&=&\displaystyle {-4\sin(12t)-3\cos(12t).}\end{array}}$ Therefore, the velocity of the object at time  $t={\frac {\pi }{8}}$ is
${\begin{array}{rcl}\displaystyle {v{\bigg (}{\frac {\pi }{8}}{\bigg )}}&=&\displaystyle {-4\sin {\bigg (}{\frac {3\pi }{2}}{\bigg )}-3\cos {\bigg (}{\frac {3\pi }{2}}{\bigg )}}\\&&\\&=&\displaystyle {-4(-1)+0}\\&&\\&=&\displaystyle {4{\text{ ft/sec}}.}\end{array}}$ position is  ${\frac {1}{4}}{\text{ ft}}.$ velocity is  $4{\text{ ft/sec}}.$ 