Yuva frequency of oscillation of a mass suspended by the spring

Dear Student
The frequency of oscillations of mass m and force constant k is
$$\begin{gathered} {\nu }_1 =\frac{1}{2\mathrm{\pi }}\sqrt{\frac{k}{m}} \\ Now \\ We know that force \left(F\right) is directly proportional to length\left(l\right) for spring \\ F=kl where k is the spring cons\tan t . \\ l\text{'}=\frac{l}{3}\left(given\right) \\ F=k\text{'}\frac{l}{3} \\ k\text{'}=\frac{3F}{l}. \\ We know k=\frac{F}{l} \\ k\text{'}=3k \\ when the spring is cut to one-third of its length, its force cons\tan t becomes 3k. \\ Then frequency of oscillation of mass m will be \\ {\nu }_2 =\frac{1}{2\mathrm{\pi }}\sqrt{\frac{3k}{m}} \\ \frac{{\nu }_2 }{{\nu }_1 } =\sqrt{3} \end{gathered}$$
So correct option is 3

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