Answer
The equation describes the motion:
$y=0.2sin(7.49t)+0.1cos(7.49t)$
The position of the block when $t=0.22s$:
$0.192m$ from the equilibrium position downward.
Work Step by Step
We can determine the required equation and phase angle as follows:
We know that:
$y=\frac{v_{o}}{ω_{n}}sinω_{n}t+y_{o}cosω_{n}t...eq(1)$
Now $K=\frac{mg}{dx}$
$K=\frac{8\times9.81}{0.175}=448.46N/m$
and $ω_{n}=\sqrt \frac{K}{m}$
$ω_{n}=\sqrt \frac{448.46}{8}=7.49 rad/s$
We plug in the known values into eq(1) to obtain the following:
$y=\frac{1.5}{7.49}sin(7.49t)+0.1cos(7.49t)$
$y=0.2sin(7.49t)+0.1cos(7.49t)...eq(2)$
To determine the position of the block when $t=0.22 s$, substitute in $eq(1)$:
$y=0.2sin(7.49t)+0.1cos(7.49t)$
$y(0.22)=0.2sin(7.49\times0.22)+0.1cos(7.49\times0.22)=0.192m$
Therefore the position of the block is $0.192m$ from the equilibrium position downward.
