An ideal spring with $k=40\frac{N}{m}$ and uncompressed length $10=0\mathrm{.}30m$ is placed on an inclined plane forming an angle of $40$ with respect to the horizontal. A m $=0\mathrm{.}45kg$ mass is attached to the uncompressed spring and let go$,$ as shown in the figure. Then, the position versus time of the resulting $5$ HM is $($consider $($:$9=9\mathrm{.}80\frac{m}{s^2}$