GATE Mechanical Engineering (ME) 2019 Shift 1 Solved Paper

natural frequency is given by $\sqrt{\frac{\mathit{m}}{{\mathit{k}}_{\mathit{e}\mathit{q}}}}$
For system I: It is a series connection For this, the equivalent ${\mathit{k}}_{\mathit{e}\mathit{q}}$ is given by
${\mathit{k}}_{\mathit{e}\mathit{q}}=\frac{{\mathit{k}}_1{\mathit{k}}_2}{{\mathit{k}}_1+{\mathit{k}}_2}=\frac{\mathit{k}\cdot \mathit{k}}{\mathit{k}+\mathit{k}}=\frac{\mathit{k}}{2}$
For system II: It is a parallel connection
For this, the equivalent ${\mathit{k}}_{\mathit{e}\mathit{q}}$ is given by
${\mathit{k}}_{\mathit{e}\mathit{q}}={\mathit{k}}_1+{\mathit{k}}_2=\mathit{k}+\mathit{k}=2\mathit{k}$
${\mathit{\omega }}_{\mathit{l}}=\sqrt{\frac{\mathit{m}}{\mathit{k}/2}}=\sqrt{\frac{2\mathit{m}}{\mathit{k}}}$
${\mathit{\omega }}_{\mathit{l}\mathit{l}}=\sqrt{\frac{\mathit{m}}{2\mathit{k}}}$
Therefore, $\frac{{\mathit{\omega }}_{\mathit{I}}}{{\mathit{\omega }}_{\mathit{l}\mathit{n}}}=\sqrt{\frac{2\mathit{m}}{\mathit{k}}\cdot \frac{2\mathit{k}}{\mathit{m}}}=\sqrt{4}=2$