JEE Mains · Physics · STD 11 - 13. oscillations
A spring stretches by \(2\) mm when it is loaded with a mass of \(200\) g. From equilibrium position the mass is further pulled down by \(2\) mm and released. The frequency associated with the system and maximum energy in the spring are __________ Hz and __________ J, respectively. (Take g \(= 10\) m/s\(^2\))
- A \(\dfrac{5\sqrt{50}}{\pi}\) and \(8 \times 10^{-3}\)
- B \(\dfrac{5\sqrt{50}}{\pi}\) and \(8\)
- C \(10\sqrt{50}\) and \(2 \times 10^{-3}\)
- D \(\dfrac{5\sqrt{50}}{\pi}\) and \(16 \times 10^{-3}\)
Answer & Solution
Correct Answer
(A) \(\dfrac{5\sqrt{50}}{\pi}\) and \(8 \times 10^{-3}\)
Step-by-step Solution
Detailed explanation
At equilibrium, the restoring force of the spring balances the weight of the mass: \(mg = kx_0\) \(k = \dfrac{mg}{x_0} = \dfrac{0.2 \times 10}{2 \times 10^{-3}} = 1000\) N/m The frequency of oscillation is given by: \(f = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}}\)…
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