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(C) The initial time period of the simple harmonic oscillator is given by $T = 2 \pi \sqrt{\frac{m}{k}}$.When a spring of length $l$ and spring consta

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$\frac{T}{\sqrt{2}}$

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(C) The initial time period of the simple harmonic oscillator is given by $T = 2 \pi \sqrt{\frac{m}{k}}$.When a spring of length $l$ and spring constant $k$ is cut into two equal parts,the length of each part becomes $l' = \frac{l}{2}$.Since the spring constant $k$ is inversely proportional to the length of the spring $(k \propto \frac{1}{l})$,the new spring constant $k'$ for each part will be $k' = 2k$.Substituting the new spring constant into the time period formula,we get the new time period $T' = 2 \pi \sqrt{\frac{m}{k'}} = 2 \pi \sqrt{\frac{m}{2k}}$.This simplifies to $T' = \frac{1}{\sqrt{2}} \left( 2 \pi \sqrt{\frac{m}{k}} \right) = \frac{T}{\sqrt{2}}$.

$A$ simple harmonic oscillator consists of a particle of mass $m$ and an ideal spring with spring constant $k$. The particle oscillates with a time period $T$. The spring is cut into two equal parts. If one part oscillates with the same particle,the time period will be

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