Answer
$4.9\times 10^{-16}Kg$
Work Step by Step
We can find the increase in the spring's mass as
$W=\frac{1}{2}Kx^2=\Delta mc^2$
This can be rearranged as:
$\Delta m=\frac{Kx^2}{2c^2}$
We plug in the known values to obtain:
$\Delta m=\frac{(584)(0.39)^2}{2(3.00\times 10^8)^2}=4.9\times 10^{-16}Kg$
