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MHT CET 2026 (18 Apr, Shift 2) · Spring Systems

Spring Systems question from MHT CET 2026 (18 Apr, Shift 2)

Q6MHT CET 2026 · 18 APR · SHIFT 2Easy

Spring Systems

Two particles A and B of equal masses are suspended from two massless springs of spring constants $K_A$ and $K_B$ respectively. The maximum velocities of the particles during oscillations are equal. The ratio of the amplitude of 'B' to that of 'A' is

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Answer: (b)

$v_{max} = A\omega = A\sqrt{\frac{K}{m}}$ .
Equal $v_{max}$ and equal $m$ : $A\sqrt{K}$ is the same, so $A \propto \frac{1}{\sqrt{K}}$ .
$\frac{A_B}{A_A} = \sqrt{\frac{K_A}{K_B}}$
So the answer is (b) $\sqrt{K_A} : \sqrt{K_B}$ .

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