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 option is 1/2.

Image Analysis and Setup:
Based on the visual representation of the spring-mass systems in the provided image:

• System I features a mass, labeled as m, connected to a rigid support via two identical springs connected in a series configuration. Each of these springs has a stiffness constant labeled as k.
• System II features the same mass m, but it is connected to a rigid support via two identical springs in a parallel configuration, with each spring having a stiffness constant labeled as k.

Let us analyze the natural frequencies of both systems step-by-step.

Step 1: Natural Frequency of System I (Series Configuration)
When two springs of stiffness constant k are connected in series, the equivalent stiffness constant $k_{eq,I}$ is given by:

$$\frac{1}{k_{eq,I}}=\frac{1}{k}+\frac{1}{k}=\frac{2}{k}$$

Taking the reciprocal, we get the equivalent stiffness for System I:

$$k_{eq,I}=\frac{k}{2}$$

The natural frequency of System I (${\omega }_I$) is:

$${\omega }_I=\sqrt{\frac{k_{eq,I}}{m}}=\sqrt{\frac{k}{2m}}$$

Step 2: Natural Frequency of System II (Parallel Configuration)
When two springs of stiffness constant k are connected in parallel, the equivalent stiffness constant $k_{eq,II}$ is the sum of the individual stiffnesses:

$$k_{eq,II}=k+k=2k$$

The natural frequency of System II (${\omega }_{II}$) is:

$${\omega }_{II}=\sqrt{\frac{k_{eq,II}}{m}}=\sqrt{\frac{2k}{m}}$$

Step 3: Calculating the Ratio of natural frequencies
To find the ratio of ${\omega }_I$ to ${\omega }_{II}$, we divide the expressions derived in Step 1 and Step 2:

$$\frac{{\omega }_I}{{\omega }_{II}}=\frac{\sqrt{\frac{k}{2m}}}{\sqrt{\frac{2k}{m}}}$$

Combining the terms under a single square root:

$$\frac{{\omega }_I}{{\omega }_{II}}=\sqrt{\frac{k}{2m}\times \frac{m}{2k}}$$

Canceling the common parameters k and m:

$$\frac{{\omega }_I}{{\omega }_{II}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$$

Therefore, the ratio of natural frequencies is 1/2.