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MHT CET 2026 (19 Apr, Shift 2) · Spring Systems

Spring Systems question from MHT CET 2026 (19 Apr, Shift 2)

Q5MHT CET 2026 · 19 APR · SHIFT 2Easy

Spring Systems

A spring force constant $180$ N/m is loaded with a mass $0.2$ kg. The amplitude of oscillations is $4$ cm. When mass comes to equilibrium position, its velocity is

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Answer: (c)

$\omega = \sqrt{k/m} = \sqrt{180/0.2} = 30\text{ rad/s}$ .
$v_{max} = A\omega = 0.04 \times 30 = 1.2\text{ m/s}$ .
So the answer is (c) $1.2$ m/s.

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