= Solution
For a <mass change in a viscously damped spring oscillator>, use linear <viscous damping> $m\ddot x+b\dot x+kx=mg$, holding the <spring constant> $k$ and <damping> coefficient $b$ fixed. Measuring $x$ from $mg/k$ removes the constant <force>. The envelope is $e^{-\gamma t}$, where
$$
\gamma=\frac b{2m}=\frac{\ln100}{30\,\mathrm s}=0.15351\,\mathrm{s^{-1}},\qquad
\omega_d=\sqrt{\frac km-\gamma^2}=\pi\,\mathrm{s^{-1}}.
$$
Doubling $m$ halves both $k/m$ and $\gamma$. Therefore
$$
\omega_{d,\rm new}^2=\frac{\omega_d^2}{2}+\frac{\gamma^2}{4},\qquad
\boxed{f_{\rm new}=0.35376\,\mathrm{Hz}\simeq0.354\,\mathrm{Hz},\qquad t_{1\%,\rm new}=60\,\mathrm s.}
$$
Neglecting <damping> in the <frequency> gives $0.5/\sqrt2=0.35355\,\mathrm{Hz}$, indistinguishable at the supplied precision. The release condition sets the <phase> and new initial <amplitude>, not these envelope ratios. A different mass-dependent drag law would require its <damping> coefficient to be specified.
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