Answer
The position at $~~t = 0~~$ is $~~-0.25~m$
Work Step by Step
We can find $\omega$:
$\omega = \sqrt{\frac{k}{m}}$
$\omega = \sqrt{\frac{100~N/m}{2.00~kg}}$
$\omega = 7.07~rad/s$
We can express the position and velocity of the block:
$x = A~cos(\omega~t+\phi)$
$v = -A~\omega~sin(\omega~t+\phi)$
Then, at $t = 1.00~s$:
$x = A~cos(7.07+\phi) = 0.129$
$v = -7.07~A~sin(7.07+\phi) = 3.415$
We can divide the second equation by the first equation to find $\phi$:
$\frac{v}{x} = -7.07~tan(7.07+\phi) = 26.47$
$tan(7.07+\phi) = -\frac{26.47}{7.07}$
$7.07+\phi = tan^{-1}~(-\frac{26.47}{7.07})$
$7.07+\phi = -1.31$
$\phi = -1.31-7.07$
$\phi = -8.38$
We can find $A$:
$x = A~cos(7.07+\phi) = 0.129$
$A~cos(7.07-8.38) = 0.129$
$A = \frac{0.129}{cos(-1.31)}$
$A = 0.50~m$
We can find the position at $t = 0$:
$x = A~cos(7.07~t+\phi)$
$x = (0.50~m)~cos[0+(-8.38)]$
$x = -0.25~m$
The position at $~~t = 0~~$ is $~~-0.25~m$