#### Answer

$\frac{1}{\pi^{2}}$

#### Work Step by Step

Since the general equation is $s(t)=a\sin w t$, we first need to compare $s(t)=a\sin w t$ with $s(t)=a\sin \sqrt \frac{k}{m}t$ to find the value of $w$.
Comparing, we find that $w=\sqrt \frac{k}{m}$. We know that $w$ and the period are linked through the formula:
Period$=\frac{2\pi}{w}$
We substitute $w=\sqrt \frac{k}{m}, k=4$ and Period$=1$ into the formula and solve:
$1=\frac{2\pi}{\sqrt \frac{k}{m}}$
$1=\frac{2\pi}{\sqrt \frac{4}{m}}$
$\sqrt \frac{4}{m}=2\pi$
$\frac{4}{m}=4\pi^{2}$
$\frac{4}{4\pi^{2}}=m$
$m=\frac{1}{\pi^{2}}$
Therefore, the mass must be $\frac{1}{\pi^{2}}$.