Two particles A and B of equal masses are suspended from two massless springs of spring constant k₁ and k₂, respectively. If the maximum velocities, during oscillation, are equal, the ratio of amplitude of A and B is (a) √((k₁)/(k₂)) (b) (k₂)/(k₁) (c) √((k₂)/(k₁)) (d) (k₁)/(k₂)

Answer: (c). vmax = A√(k/m); equal vmax and m give A ∝ 1/√(k). (AA)/(AB) = √((k₂)/(k₁)) So the answer is (c) √((k₂)/(k₁)).

Two particles A and B of equal masses are suspended from two massless… | JEE Main 2003 (AIEEE)

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Spring Systems

Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constant $k_1$ and $k_2$ , respectively. If the maximum velocities, during oscillation, are equal, the ratio of amplitude of $A$ and $B$ is

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Answer: (c)

$v_{max} = A\sqrt{\frac{k}{m}}$ ; equal $v_{max}$ and $m$ give $A \propto \frac{1}{\sqrt{k}}$ .
$\frac{A_A}{A_B} = \sqrt{\frac{k_2}{k_1}}$
So the answer is (c) $\sqrt{\frac{k_2}{k_1}}$ .

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