A spring executes S.H.M. with mass 1 kg attached to it. The force constant of the spring is 4 N/m. If at any instant its velocity is 20 cm/s, the displacement at that instant is (Amplitude of S.H.M. is 0.4 m) (a) √(0.11) m (b) √(0.15) m (c) √(0.17) m (d) √(0.19) m

Answer: (b). ω = √(k/m) = √(4/1) = 2 rad/s v = ω√(A² - x²): 0.2 = 2√(0.16 - x²) 0.01 = 0.16 - x² ⇒ x² = 0.15, x = √(0.15) m So the answer is (b) √(0.15) m.

A spring executes S.H.M. with mass 1 kg attached to it. The force… | MHT CET 2025 (26 Apr, Shift 1)

Q26MHT CET 2025 · 26 APR · SHIFT 1Medium

Spring Systems

A spring executes S.H.M. with mass 1 kg attached to it. The force constant of the spring is $4\text{ N/m}$ . If at any instant its velocity is $20\text{ cm/s}$ , the displacement at that instant is (Amplitude of S.H.M. is 0.4 m)

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Answer: (b)

$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{4}{1}} = 2\text{ rad/s}$
$v = \omega\sqrt{A^2 - x^2}$ : $0.2 = 2\sqrt{0.16 - x^2}$
$0.01 = 0.16 - x^2 \Rightarrow x^2 = 0.15$ , $x = \sqrt{0.15}\text{ m}$
So the answer is (b) $\sqrt{0.15}$ m.

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Spring Systems question from MHT CET 2025 (26 Apr, Shift 1)

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