A mass m is suspended from a spring of length | vedclass

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(D) The frequency of a spring-mass system is given by $f = \frac{1}{2\pi} \sqrt{\frac{K}{m}}$.Initially,the frequency is $f_1 = \frac{1}{2\pi} \sqrt{\

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$f_2 = \sqrt{2} f_1$

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(D) The frequency of a spring-mass system is given by $f = \frac{1}{2\pi} \sqrt{\frac{K}{m}}$.Initially,the frequency is $f_1 = \frac{1}{2\pi} \sqrt{\frac{K}{m}}$.When a spring of force constant $K$ and length $l$ is cut into two equal parts,the force constant of each part becomes $K' = 2K$ because the force constant is inversely proportional to the length $(K \propto \frac{1}{l})$.When the same mass $m$ is suspended from one of these parts,the new frequency is $f_2 = \frac{1}{2\pi} \sqrt{\frac{2K}{m}}$.Comparing the two frequencies,we get $f_2 = \sqrt{2} \left( \frac{1}{2\pi} \sqrt{\frac{K}{m}} \right) = \sqrt{2} f_1$.Therefore,the correct relation is $f_2 = \sqrt{2} f_1$.

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