Question:

A spring stretches by \(2\,\text{mm}\) when it is loaded with a mass of \(200\,\text{g}\). From equilibrium position the mass is further pulled down by \(2\,\text{mm}\) and released. The frequency associated with the system and the maximum energy in the spring are _____ Hz and _____ J, respectively. (Take \(g=10\,\text{m/s}^2\)).

Updated On: Apr 12, 2026
  • \( \frac{5\sqrt{50}}{\pi} \) and \(8\times10^{-3}\)
  • \( \frac{5\sqrt{50}}{\pi} \) and \(8\)
  • \( \frac{10\sqrt{50}}{\pi} \) and \(2\times10^{-3}\)
  • \( \frac{5\sqrt{50}}{\pi} \) and \(16\times10^{-3}\)
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The Correct Option is A

Solution and Explanation

Concept: For a spring system: \[ mg=kx \] and frequency of oscillation is \[ f=\frac{1}{2\pi}\sqrt{\frac{k}{m}} \] Maximum energy in SHM: \[ E=\frac12 kA^2 \] Step 1: {Find spring constant.} Given \[ m=0.2\,\text{kg} \] \[ x=2\,\text{mm}=2\times10^{-3}\,\text{m} \] Using \(mg=kx\): \[ k=\frac{mg}{x} \] \[ k=\frac{0.2\times10}{2\times10^{-3}} \] \[ k=1000\,\text{N/m} \] Step 2: {Find frequency.} \[ f=\frac{1}{2\pi}\sqrt{\frac{1000}{0.2}} \] \[ =\frac{1}{2\pi}\sqrt{5000} \] \[ =\frac{5\sqrt{50}}{\pi} \] Step 3: {Find maximum energy.} Amplitude: \[ A=2\,\text{mm}=2\times10^{-3}\,\text{m} \] \[ E=\frac12 (1000)(2\times10^{-3})^2 \] \[ =8\times10^{-3}\,\text{J} \]
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