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 )

frac 5√(50) π and 8times10 -3

frac 5√(50) π and 16times10 -3

Physics · Simple Harmonic Motion

JEE Main 2026 — 6 April, Evening Shift — Question 9

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

Option A:
$\frac{5\sqrt{50}}{\pi}$ and $8\times10^{-3}$
Correct
Option B:
$\frac{5\sqrt{50}}{\pi}$ and 8
Option C:
$10\sqrt{50}$ and $2\times10^{-3}$
Option D:
$\frac{5\sqrt{50}}{\pi}$ and $16\times10^{-3}$

Answer: A

Step-by-step solution

Spring constant $k = mg/x = 0.2\times10/0.002=1000$ N/m. Frequency $f = \frac{1}{2\pi}\sqrt{k/m} = \frac{1}{2\pi}\sqrt{1000/0.2} = \frac{\sqrt{5000}}{2\pi} = \frac{50\sqrt{2}}{2\pi} = \frac{25\sqrt{2}}{\pi} = \frac{5\sqrt{50}}{\pi}$ . Max energy = $\frac12 k A^2$ with A=0.004 m, $E=0.5\times1000\times16\times10^{-6}=8\times10^{-3}$ J.

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 6 April, Evening Shift
Subject: Physics
Chapter: Simple Harmonic Motion
Topic: Linear SHM and Spring-Pulley-Block Systems