Answer
$x = 0.26~cos(20~t-\frac{\pi}{2})$
Work Step by Step
In part (b), we found that the amplitude is $A = 0.26~m$
We can find $\omega$:
$\omega = \sqrt{\frac{k}{m}}$
$\omega = \sqrt{\frac{480~N/m}{1.2~kg}}$
$\omega = 20~rad/s$
We can write an expression for $x$ as a function of time:
$x = A~cos(\omega~t+\phi)$
Since the block passes through $x = 0$ at time $t=0$ and it is moving in the positive direction, then $\phi = -\frac{\pi}{2}$
We can write an expression for $x$ as a function of time:
$x = A~cos(\omega~t+\phi)$
$x = 0.26~cos(20~t-\frac{\pi}{2})$
