#### Answer

Amplitude= 0.5
Period$=4.44$ seconds
Frequency$=0.225$ cycles per second

#### Work Step by Step

In the equation $s(t)=a\sin \sqrt \frac{k}{m}t$, we substitute the values of $k$ and $m$:
$s(t)=a\sin \sqrt \frac{k}{m}t$
$s(t)=a\sin \sqrt \frac{2}{1}t$
$s(t)=a\sin \sqrt{2} t$
Since we know that the spring is stretched 0.5 ft and then released, this means that the amplitude is 0.5. Next, to find the period, we need to compare the equation above with the standard equation $s(t)=a\sin w t$.
Comparing the two equations, we find that $w=\sqrt 2$. Therefore, the period can be found through the formula:
Period$=\frac{2\pi}{w}$
Period$=\frac{2\pi}{\sqrt 2}$
Period$=4.44$ seconds
Frequency is the reciprocal of period. Therefore,
Frequency$=\frac{1}{4.44}=0.225$ cycles per second