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(C) The frequency of oscillation of a mass $m$ attached to a spring is given by $v = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$, where $k$ is the spring const

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$\sqrt{2}$

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(C) The frequency of oscillation of a mass $m$ attached to a spring is given by $v = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$, where $k$ is the spring constant.It is known that the spring constant $k$ is inversely proportional to the length $L$ of the spring, i.e.,$k \propto 1/L$.When the length of the spring is cut to half $(L' = L/2)$, the new spring constant $k'$ becomes $k' = k \cdot (L/L') = k \cdot (L / (L/2)) = 2k$.The new frequency $v_2$ is given by $v_2 = \frac{1}{2\pi} \sqrt{\frac{k'}{m}} = \frac{1}{2\pi} \sqrt{\frac{2k}{m}}$.This can be written as $v_2 = \sqrt{2} \cdot \left( \frac{1}{2\pi} \sqrt{\frac{k}{m}} \right) = \sqrt{2} v_1$.Therefore, the ratio $v_2/v_1 = \sqrt{2}$.

The frequency of oscillation of a mass $m$ suspended by a spring is $v_1$. If the length of the spring is cut to half, the same mass oscillates with frequency $v_2$. The value of $v_2/v_1$ is . . . . . . .

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