The natural frequencies corresponding to the spring-mass systems I and II are ω_I and ω_II, respectively. The ratio ω_I / ω_II is

The correct answer is 1/2.

1. Analysis of System I (Series Connection):
From the image, System I consists of two springs, each of stiffness $k$, connected in series with a mass $M$.
For a series connection of two springs, the equivalent spring stiffness $k_{\text{eq1}}$ is calculated as:
$\frac{1}{k_{\text{eq1}}}=\frac{1}{k}+\frac{1}{k}=\frac{2}{k}$
Taking the reciprocal, we get:
$k_{\text{eq1}}=\frac{k}{2}$
The natural frequency of System I (${\omega }_{\text{I}}$) is given by:
${\omega }_{\text{I}}=\sqrt{\frac{k_{\text{eq1}}}{M}}=\sqrt{\frac{k}{2M}}$

2. Analysis of System II (Parallel Connection):
From the image, System II consists of two springs, each of stiffness $k$, connected in parallel to a mass $M$.
For a parallel connection of two springs, the equivalent spring stiffness $k_{\text{eq2}}$ is calculated as:
$k_{\text{eq2}}=k+k=2k$
The natural frequency of System II (${\omega }_{\text{II}}$) is given by:
${\omega }_{\text{II}}=\sqrt{\frac{k_{\text{eq2}}}{M}}=\sqrt{\frac{2k}{M}}$

3. Ratio of the Natural Frequencies:
Now, we calculate the ratio of ${\omega }_{\text{I}}$ to ${\omega }_{\text{II}}$:
$\frac{{\omega }_{\text{I}}}{{\omega }_{\text{II}}}=\frac{\sqrt{\frac{k}{2M}}}{\sqrt{\frac{2k}{M}}}=\sqrt{\frac{k/\left(2M\right)}{2k/M}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$
Therefore, the ratio of natural frequencies is 1/2.