Answer
(a) The pattern traced on the paper will be a vertical line that has a length of $0.24~m$.
(b) $y(t) = (0.12~cm)~cos~[~(9.0~rad/s)~t~]$
On the graph below, we can see the pattern that would be traced on the paper.
Work Step by Step
(a) At the equilibrium point, the upward force of the spring is equal in magnitude to the object's weight. We can find the extension of the spring at the equilibrium point:
$kd = mg$
$d = \frac{mg}{k}$
$d = \frac{(0.306~kg)(9.80~m/s^2)}{25~N/m}$
$d = 0.12~m$
Since the extension distance of the spring at the equilibrium point is $0.12~m$, the lowest point in the motion will be twice this distance. Therefore, the pattern traced on the paper will be a vertical line that has a length of $0.24~m$.
(b) We can find $\omega$:
$\omega = \sqrt{\frac{k}{m}}$
$\omega = \sqrt{\frac{25~N/m}{0.306~kg}}$
$\omega = 9.0~rad/s$
In general: $y(t) = A~cos(\omega~t+\phi)$
In this case: $y(t) = (0.12~cm)~cos~[~(9.0~rad/s)~t~]$
On the graph below, we can see the pattern that would be traced on the paper.
