Answer
(a) $y=16e^{-0.72t}cos(2.8\pi t)$
(b) see graph
(c) $0.012cm$
Work Step by Step
At time t = 0 the displacement is at a maximum, we should use a $cos$ function.
(a) Given $k=16,c=0.72,f=1.4,\omega=2\pi f =2.8\pi$. the function becomes
$y=16e^{-0.72t}cos(2.8\pi t)$
(b) see graph
(c) when $t=10s$, $y=16e^{-7.2}cos(28\pi)=0.012cm$
