Question 299: Oscillations

Question

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 $$\pi^2=10$$ )

Accepted Answer

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,
$$\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}} \
& 	herefore \quad \mathrm{T}^2=4 \pi^2 	imes \frac{\mathrm{m}}{2 \mathrm{k}} \
& 	herefore \quad \mathrm{k}=4 \pi^2 	imes \frac{\mathrm{m}}{2 \mathrm{~T}^2}=4 	imes 10 	imes \frac{12}{2 	imes 4}=60 \mathrm{~N} / \mathrm{m}
\end{aligned}$$

Answer

230 Nm$$^{-1}$$

Answer

120 Nm$$^{-1}$$

Answer

60 Nm$$^{-1}$$

Answer

30 Nm$$^{-1}$$

Oscillations

A uniform circular disc of mass $12 \mathrm{~kg} is held by two identical springs. When the disc is slightly pressed down and released, it executes S.H.M. of period 2 second. The force constant of each spring is (nearly) (Take \pi^2=10$ )

Choose the correct answer: