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A mass $\mathrm{m}$ 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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Verified Answer
The correct answer is:
$\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}+\pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}$.
When the spring undergoes displacement in the downward direction it completes one half oscillation while it completes another halfoscillation in the upward direction. The total time period is:
$$
\mathrm{T}=\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}+\pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}
$$
$$
\mathrm{T}=\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}+\pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}
$$
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