Suppose that an object is attached to a coiled spring. It is pulled down a distance of 8 cm from is equilibrium position and then released.
Therefore, a=8 cm.
It takes 1/3 second to complete one oscillation.
Therefore, t=1/3
we know that,
[tex]S(t)=-a\cos wt[/tex]So, we have
[tex]S(t)=-8\cos wt\ldots\ldots(1)[/tex]first we need to find w :
The formula to find w is,
[tex]\begin{gathered} w=\frac{2\pi}{t} \\ =\frac{2\pi}{\frac{1}{3}} \\ =\frac{6\pi}{3} \end{gathered}[/tex]Therefore, the equation (1) becomes,
[tex]S(t)=-8\cos \frac{6\pi}{3}t[/tex]Next to find the position of the object after 1/2 second.
substituting t=1/2 in the above equation we get,
[tex]\begin{gathered} S(t)=-8\cos (\frac{6\pi}{3}.\frac{1}{2}) \\ =-8\cos (\frac{3\pi}{3}) \\ =-8\cos \pi \\ =-8(-1)_{} \\ =8 \end{gathered}[/tex]Hence, the position of the object after 1/2 second is 8 cm.
