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(B) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$,where $m$ is the mass and $k$ is the spring constant.From this,w

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$1: \sqrt{2}$

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(B) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$,where $m$ is the mass and $k$ is the spring constant.From this,we see that $T \propto \frac{1}{\sqrt{k}}$.Therefore,the ratio is $\frac{T_2}{T_1} = \sqrt{\frac{k_1}{k_2}}$.When a spring of constant $k$ is cut into two equal halves,the spring constant of each half becomes $k' = 2k$.In the first case,the spring constant is $k_1 = k$.In the second case,the mass is suspended from one half,so the new spring constant is $k_2 = 2k$.Substituting these values into the ratio formula:$\frac{T_2}{T_1} = \sqrt{\frac{k}{2k}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

$A$ spring has a certain mass suspended from it and its period of vertical oscillations is $T_1$. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now $T_2$. The ratio of $T_2 / T_1$ is

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