A mass m is attached to two identical springs having stiffness k as shown in figure below. The natural frequency of the vibrating system is

Correct option is B

​$$\begin{aligned}&\text{When a mass is connected between two springs, and both oppose displacement:} \\&k_{\text{eq}} = k_{\text{top}} + k_{\text{bottom}} = k + k = 2k \\[1em]&\text{\underline{Natural Frequency Formula:}} \\&\omega_n = \sqrt{\frac{k_{\text{eq}}}{m}} = \sqrt{\frac{2k}{m}} \\[1em]&\text{\ \textbf{Final Answer:}} \\&\boxed{\omega_n = \sqrt{\frac{2k}{m}}} \\[1em]&\text{\underline{Information Booster:}} \\&1. \ \text{Both springs act together, increasing total stiffness.} \\&2. \ \text{If in series: } k_{\text{eq}} = \frac{k}{2}, \text{ so frequency would be lower.} \\&3. \ \text{Natural frequency units: rad/s} \\&4. \ \text{Frequency in Hz:} \\&f_n = \frac{\omega_n}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{2k}{m}}\end{aligned}$$​​