Problem 12.9

12.9. a)
We know that the spring constant is given by
$\omega = \sqrt{\frac{k}{m}} ~=~ 4 s^{-1}$
therefore the equation for the position as a function of time is
$x(t) = 10 \sin 4t$
and likewise,
$v(t) = 40 \cos 4t$
$a(t) = -160 \sin 4t$
and v_max = 40 cm/s and a_max = 160 cm/s.

b)
When x=6cm, we can solve for t
$t=\frac{1}{4} \sin^{-1} (\frac{6}{10})$
or t= .161 sec
then, $v = 40 \cos (4)(.161) ~=~ 32 cm/s$
and $a = -160 \sin (4)(.161) ~=~ -96 cm/s^{2}$

c)
When x=0, t=0, and so when x=8cm, t=.232
$\Delta t = .232 sec$.