Two particles A and B of equal masses are suspended from two massless springs of spring constants K1 and K2 respectively. If the maximum velocities during oscillations are equal, the ratio of the amplitude of A and B is
Solution
$\because$ ${V_{\max }} = A\omega$<br><br>Given, ${\omega _1}{A_1} = {\omega _2}{A_2}$<br><br>We know that $\omega = \sqrt {{K \over m}}$<br><br>$\therefore$ $\sqrt {{{{k_1}} \over m}} {A_1} = \sqrt {{{{k_2}} \over m}} {A_2}$<br><br>$\Rightarrow$ ${{{A_1}} \over {{A_2}}} = \sqrt {{{{k_2}} \over {{k_1}}}}$
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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