thanks skeeter
hey skeeter i dont actually need the graph, but the working out to get to that point. If you know how?
plus when i plug that equation into my ti-83 the graph doesnt look the same.
your using a mac program for that right?
my mistake ... I misread the 5.2 degrees "either side" as the total oscillation.
General form for the equation that models this motion will be
$\displaystyle
y = A\cos(\omega t + \delta)
$
period = $\displaystyle T = 30 \, sec \, = \frac{2\pi}{\omega}$
$\displaystyle
\omega = \frac{\pi}{15}
$
$\displaystyle
A = -5.2
$
at this point ...
$\displaystyle
y = -5.2\cos\left(\frac{\pi}{15} t + \delta\right)
$
at $\displaystyle t = 0$ , $\displaystyle y = -2.6$
$\displaystyle
-2.6 = -5.2\cos\left(\frac{\pi}{15} (0) + \delta\right)
$
$\displaystyle
\frac{1}{2} = \cos\left(\delta\right)
$
$\displaystyle
\delta = \frac{\pi}{3}
$
so ...
$\displaystyle
y = -5.2\cos\left(\frac{\pi}{15} t + \frac{\pi}{3}\right)
$
make sure your calculator is in radian mode when you graph it.