A mass m₁ performs S.H.M. with amplitude A which is connected to horizontal spring. While mass m₁ passing through mean position, another mass m₂ ( m₂ < m₁) is placed on it so that both masses move together with amplitude A₁. The ratio of A/A₁ is (a) [(m₁+m₂)/(m₁)]1/2 (b) [(m₁)/(m₁+m₂)]1/2 (c) (m₁)/(m₁+m₂) (d) (m₂)/(m₁+m₂)

Answer: (a). Momentum conserved: m₁v = (m₁+m₂)v'. A = (v)/(ω) with ω = √(k/m), so A ∝ (v)/(√(k/m)) = (mv)/(√(km)) = (p)/(√(km)). Same p: A/A₁ = √((m₁+m₂)/(m₁)) So the answer is (a).

A mass m₁ performs S.H.M. with amplitude A which is connected to… | MHT CET 2026 (16 Apr, Shift 2)

Q21MHT CET 2026 · 16 APR · SHIFT 2Medium

Spring Systems

A mass $m_1$ performs S.H.M. with amplitude A which is connected to horizontal spring. While mass $m_1$ passing through mean position, another mass $m_2$ ( $m_2 < m_1$ ) is placed on it so that both masses move together with amplitude $A_1$ . The ratio of $A/A_1$ is

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Answer: (a)

Momentum conserved: $m_1v = (m_1+m_2)v'$ .
$A = \frac{v}{\omega}$ with $\omega = \sqrt{k/m}$ , so $A \propto \frac{v}{\sqrt{k/m}} = \frac{mv}{\sqrt{km}} = \frac{p}{\sqrt{km}}$ .
Same $p$ : $\frac{A}{A_1} = \sqrt{\frac{m_1+m_2}{m_1}}$
So the answer is (a).

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Spring Systems question from MHT CET 2026 (16 Apr, Shift 2)

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