Frequency of Oscillation for Cut Spring Mass System

To find the ratio of the frequencies of oscillation, we analyze how the spring constant depends on the length of the spring.

The frequency of oscillation $\nu$ of a mass $m$ attached to a spring with spring constant $k$ is given by: $\nu = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

For a given spring material and cross-section, the spring constant $k$ is inversely proportional to its original length $L$: $k \propto \frac{1}{L} \implies k \cdot L = \text{constant}$

Let the initial length of the spring be $L_1 = L$ and its spring constant be $k_1$. The initial frequency of oscillation is: $\nu_1 = \frac{1}{2\pi} \sqrt{\frac{k_1}{m}}$

When the spring is cut to half of its original length, its new length becomes $L_2 = \frac{L}{2}$. Using the inverse proportionality relation: $k_2 L_2 = k_1 L_1$ $k_2 \left(\frac{L}{2}\right) = k_1 L \implies k_2 = 2k_1$

The new frequency of oscillation $\nu_2$ with the same mass $m$ is: $\nu_2 = \frac{1}{2\pi} \sqrt{\frac{k_2}{m}} = \frac{1}{2\pi} \sqrt{\frac{2k_1}{m}}$

Taking the ratio of the two frequencies: $\frac{\nu_2}{\nu_1} = \frac{\frac{1}{2\pi} \sqrt{\frac{2k_1}{m}}}{\frac{1}{2\pi} \sqrt{\frac{k_1}{m}}} = \sqrt{\frac{2k_1}{k_1}} = \sqrt{2}$

Thus, the value of $\frac{\nu_2}{\nu_1}$ is $\sqrt{2}$, which corresponds to Option C.