Q104 / 125 · BITSAT 2011 · Physics
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
- A$2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$
- B$\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}+\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k} / 2}}$
- C$\pi \sqrt{\frac{\mathrm{m}}{3 \mathrm{k} / 2}}$
- Correct answerD$\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}+\pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}$.
Solution
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:
$$
\mathrm{T}=\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}+\pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}
$$
Hence, the correct answer is option: 'D'.