Spring Calculation - find mass A spring with an $-kg$ mass and a damping constant $9$ can be held stretched $2.5 \text{ meters}$ beyond its natural length by a force of $7.5 \text{ Newtons}$. If the spring is stretched $5 \text{ meters}$ beyond its natural length and then released with zero velocity, find the mass that would produce critical damping. 


My work:

The restoring force is $-kx$.  Then

$$7.5 = -k(2.5) \\
-\frac{7.5}{2.5} = k \\ 
ma = -\frac{7.5x}{2.5} \\
my’’ + 9y’ + -3y = 0,\quad y(0) = 2.5, y(5) = 0 \\
\frac{-9 \pm \sqrt{81 + 4(m)(3)}}{2m} \\
-\frac{9}{2m} \pm \frac{\sqrt{81+12m}}{2m} \\
y = Ae^{-(9/2)x}\cos\left(\frac{\sqrt{81+12m}}{2m}x\right) + Be^{-(9/2)x}\sin\left(\frac{\sqrt{81+12m}}{2m}x\right) \\
2.5 = A + B\cdot 0 \\
0 = (2.5)e^{-45/2}\cos\left(\sqrt{81+12m}\frac{5}{2m}\right) + Be^{-45/2}\sin\left(\sqrt{81+12m}\frac{5}{2m}\right)$$

Any help would be appreciated
1) find the $k$

2)$\zeta=c/(2\sqrt{k/m})=1 $ for critical damping. So solve for $m$ as you alreaday know $c$ and $k$.
Critical damping occurs when $(\gamma^2)-4mk=0$
Therefore,
$$(9^2)-4m(7.5/2.5)=0$$
$$81=12m$$
$$m=6.75 kg$$
Bob Shannon also got the same answer, but I wanted to present it this way because I thought it might be more clear to some.