Question

Sol. [Key Concept: $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$; halving the spring doubles $k$, so $\frac{v_{2}}{v_{1}} = \sqrt{2}$.]

For a spring, $k\mathcal{l} =$ constant

$${k_{1}\mathcal{l} = k_{2}\frac{\mathcal{l}}{2} \Rightarrow k_{2} = 2k_{1} }{f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} }{\Rightarrow \frac{f_{1}}{f_{2}} = \sqrt{\frac{k_{1}}{k_{2}}} }{\Rightarrow \frac{f_{1}}{f_{2}} = \sqrt{\frac{1}{2}} \Rightarrow \frac{f_{2}}{f_{1}} = \sqrt{2} }$$