given: \(f_{1} ; f_{2}=4: 1\) i.e.
\(\frac{f_{1}}{f_{2}}=\frac{4}{1}\); required: \(m_{1}: m_{2}=\) ?
correlating equation
$$
f=\frac{1}{2 l} \sqrt{\frac{T}{m}} \Rightarrow f \alpha \frac{1}{\sqrt{m}}
$$
\begin{aligned}
f_{1} \sqrt{m_{1}} &=f_{2} \sqrt{m_{2}} \\
\frac{f_{1}}{f_{2}} &=\sqrt{\frac{m_{2}}{m_{1}}} ; \frac{f_{1}^{2}}{f_{2}^{2}}=\frac{m_{2}}{m_{1}}=\frac{4^{2}}{1^{2}} \\
\frac{m_{2}}{m_{1}} &=\frac{16}{1} ; m_{1}: m_{2}=1: 16
\end{aligned}