The time period of simple harmonic motion of mass M in the given figure is π√((α M)/(5k)), where the value of α is _________. Numerical answer type. Work it out, then open the answer.

Answer: 12. Right branch: two k in parallel give 2k, in series with k: (2k· k)/(3k) = 2k/3. This branch is in parallel with the left spring k: keff = k + 2k/3 = 5k/3. T = 2π√(3M/5k) = π√(12M/5k) α = 12

The time period of simple harmonic motion of mass M in the given… | JEE Main 2024 (31 Jan, Shift 2)

Q44JEE Main 2024 · 31 JAN · SHIFT 2Medium

Spring Systems

The time period of simple harmonic motion of mass $M$ in the given figure is $\pi\sqrt{\frac{\alpha M}{5k}}$ , where the value of $\alpha$ is _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 12

Right branch: two $k$ in parallel give $2k$ , in series with $k$ : $\frac{2k\cdot k}{3k} = \frac{2k}{3}$ .
This branch is in parallel with the left spring $k$ : $k_{eff} = k + \frac{2k}{3} = \frac{5k}{3}$ .
$T = 2\pi\sqrt{\frac{3M}{5k}} = \pi\sqrt{\frac{12M}{5k}}$
$\alpha = 12$

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