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MHT CET 2026 (13 Apr, Shift 1) · Spring Systems

Spring Systems question from MHT CET 2026 (13 Apr, Shift 1)

Q11MHT CET 2026 · 13 APR · SHIFT 1Medium

Spring Systems

A spring executes S.H.M. with mass 10 kg attached to it. The force constant of spring is 10 N/m. If at any instant its velocity is 40 cm/s, the displacement at that instant is

(Amplitude of S.H.M. is 0.5 m)

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Answer: (b)

  1. ω=km=1010=1 rad/s\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{10}{10}} = 1\text{ rad/s}
  2. v=ωA2−x2v = \omega\sqrt{A^2 - x^2}: 0.4=1×0.25−x20.4 = 1 \times \sqrt{0.25 - x^2}
  3. x2=0.25−0.16=0.09⇒x=0.3 mx^2 = 0.25 - 0.16 = 0.09 \Rightarrow x = 0.3\text{ m}
  4. So the answer is (b) 0.30.3 m.
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