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

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(D) When the mass $m$ moves upward,it is attached only to the top spring with force constant $k$. The time taken for this half-oscillation is $t_1 = \

Vedclass · Accepted Answer

$\pi \sqrt{\frac{ m }{ k }}+\pi \sqrt{\frac{ m }{2 k }}$

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(D) When the mass $m$ moves upward,it is attached only to the top spring with force constant $k$. The time taken for this half-oscillation is $t_1 = \pi \sqrt{\frac{m}{k}}$.When the mass $m$ moves downward,it compresses both the top spring and the bottom spring. Since both springs have force constant $k$ and are in parallel,the effective spring constant is $k_{eff} = k + k = 2k$. The time taken for this half-oscillation is $t_2 = \pi \sqrt{\frac{m}{2k}}$.The total time period of one full oscillation is the sum of the times for the two half-oscillations:$T = t_1 + t_2 = \pi \sqrt{\frac{m}{k}} + \pi \sqrt{\frac{m}{2k}}$.

$A$ mass $m$ is suspended from a spring of force constant $k$ and just touches another identical spring fixed to the floor as shown in the figure. The time period of small oscillations is

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