A spring is stretched $190$ mm by a $13$ kg block.

Part A

If the block is displaced $80$ mm downward from its equilibrium position and given a downward velocity of $0\mathrm{.}80($m$)/($s$),$ determine the equation which describes the motion. Assume that positive displacement is

downward.

y$=0\mathrm{.}111$sin$7\mathrm{.}19$t$+8\mathrm{.}00\backslash$times $10^(-2)$cos$7\mathrm{.}19$t

y$=0\mathrm{.}111$sin$7\mathrm{.}19$t$-0\mathrm{.}190$cos$7\mathrm{.}19$t

y$=0\mathrm{.}111$cos$7\mathrm{.}19$t$-8\mathrm{.}00\backslash$times $10^(-2)$sin$7\mathrm{.}19$t

y$=-0\mathrm{.}111$sin$7\mathrm{.}19$t$+8\mathrm{.}00\backslash$times $10^(-2)$cos$7\mathrm{.}19$t

Part B

What is the phase angle?

Express your answer with the appropriate units.

$\backslash$phi $=$