= Solution
For the same massless <spring> of <spring constant> $k$, <Hooke's law> gives the vertical <simple harmonic motion> about the displaced <equilibrium> position:
$$
\nu_i=\frac1{2\pi}\sqrt{\frac{k}{m_i}},\qquad
R=\frac{m_1}{m_2}=\left(\frac{\nu_2}{\nu_1}\right)^2.
$$
Gravity shifts the <equilibrium> extension but does not change this <frequency>. Use the measured <sample means> in the ratio estimate. For small measurement errors, <logarithmic differentiation> gives
$$
\frac{dR}{R}=2\frac{d\nu_2}{\nu_2}-2\frac{d\nu_1}{\nu_1}.
$$
Assuming the two frequency estimates have <independent> errors, their <variances> add. Writing $\epsilon=\delta\nu/\overline\nu$ for their common fractional <standard uncertainty>, <monomial uncertainty propagation> yields
$$
\left(\frac{\sigma(R)}R\right)^2\simeq4\epsilon^2+4\epsilon^2,
\qquad
\boxed{\frac{\sigma(R)}R\simeq\sqrt8\,\frac{\delta\nu}{\overline\nu}.}
$$
This is a first-order uncertainty formula, rather than an exact identity for an arbitrary noisy ratio. It assumes that errors in the two measured <frequencies> have zero <covariance>.
There is a statistical distinction in the wording: the <standard deviation> of individual repetitions is not the <standard error> of their <sample mean>. If the supplied $\sigma(\nu_i)$ describes individual trials, the frequency means have errors $\sigma(\nu_i)/\sqrt n$, and the ratio formed from those means instead satisfies
$$
\boxed{\frac{\sigma(R_{\rm means})}{R_{\rm means}}\simeq
\sqrt{\frac8n}\,\frac{\delta\nu}{\overline\nu}.}
$$
Thus the printed formula describes either a single-pair ratio or a convention in which $\delta\nu$ already denotes the uncertainty of each averaged <frequency>. Under either consistent interpretation, <independent> repeated measurements reduce statistical error in proportion to $n^{-1/2}$. Halving it requires \b[$4n$ trials in total for each mass, or $3n$ additional trials]. Repetition alone does not reduce a fixed <systematic error>.
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