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MHT CET 2026 (20 May, Shift 1) · Spring Systems

Spring Systems question from MHT CET 2026 (20 May, Shift 1)

Q3MHT CET 2026 · 20 MAY · SHIFT 1Easy

Spring Systems

A spring of force constant $180$ N/m is loaded with a mass $0.20$ kg. The amplitude of oscillations is $4$ cm. When mass comes to equillibrium position, its velocity in m/s is

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Answer: (b)

$\omega = \sqrt{k/m} = \sqrt{180/0.20} = 30\text{ rad/s}$ .
At equilibrium $v_{max} = A\omega = 0.04 \times 30 = 1.2\text{ m/s}$ .
So the answer is (b) $1.2$ .

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