# 009A Sample Midterm 1, Problem 5

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

${\displaystyle y={\frac {1}{3}}\cos(12t)-{\frac {1}{4}}\sin(12t)}$

where  ${\displaystyle y}$  is measured in feet and  ${\displaystyle t}$  is the time in seconds.

Determine the position and velocity of the object when  ${\displaystyle t={\frac {\pi }{8}}.}$

Foundations:
What is the relationship between position  ${\displaystyle s(t)}$  and velocity  ${\displaystyle v(t)}$  of an object?
${\displaystyle v(t)=s'(t)}$

Solution:

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

position is  ${\displaystyle {\frac {1}{4}}{\text{ foot}}.}$
velocity is  ${\displaystyle 4{\text{ feet/second}}.}$