## Trigonometry (11th Edition) Clone

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