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A uniform circular disc of mass $12 \mathrm{~kg}$ is held by two identical springs. When the disc is slightly pressed down and released, it executes S.H.M. of period 2 second. The force constant of each spring is (nearly) (Take $\pi^2=10$ )
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Verified Answer
The correct answer is:
$60 \mathrm{Nm}^{-1}$
The two springs are connected in parallel. So, the effective spring constant is, $\mathrm{k}_{\mathrm{eff}}=2 \mathrm{k}$
Time period of the spring system is,
$\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}} \\
\therefore \quad \mathrm{T}^2 & =4 \pi^2 \times \frac{\mathrm{m}}{2 \mathrm{k}} \\
\therefore \quad \mathrm{k} & =4 \pi^2 \times \frac{\mathrm{m}}{2 \mathrm{~T}^2}=4 \times 10 \times \frac{12}{2 \times 4}=60 \mathrm{~N} / \mathrm{m}
\end{aligned}$
Time period of the spring system is,
$\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}} \\
\therefore \quad \mathrm{T}^2 & =4 \pi^2 \times \frac{\mathrm{m}}{2 \mathrm{k}} \\
\therefore \quad \mathrm{k} & =4 \pi^2 \times \frac{\mathrm{m}}{2 \mathrm{~T}^2}=4 \times 10 \times \frac{12}{2 \times 4}=60 \mathrm{~N} / \mathrm{m}
\end{aligned}$
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