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JEE Main 2026 (6 Apr, Shift 2) · Spring Systems

Spring Systems question from JEE Main 2026 (6 Apr, Shift 2)

Q14JEE Main 2026 · 6 APR · SHIFT 2Medium

Spring Systems

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/s2g = 10\text{ m/s}^2)

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Answer: (a)

  1. k=mgx=0.2×102×10−3=1000k = \frac{mg}{x} = \frac{0.2\times10}{2\times10^{-3}} = 1000 N/m
  2. f=12πkm=50002π=550πf = \frac{1}{2\pi}\sqrt{\frac{k}{m}} = \frac{\sqrt{5000}}{2\pi} = \frac{5\sqrt{50}}{\pi} Hz
  3. Lowest point: total extension =2+2=4= 2 + 2 = 4 mm.
  4. Umax=12(1000)(4×10−3)2=8×10−3U_{max} = \frac{1}{2}(1000)(4\times10^{-3})^2 = 8\times10^{-3} J
  5. So the answer is (a).
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