JEE MAIN - Physics (2021 - 17th March Evening Shift - No. 11)
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
$${{{K_1}} \over {{K_2}}}$$
$$\sqrt {{{{K_1}} \over {{K_2}}}} $$
$${{{K_2}} \over {{K_1}}}$$
$$\sqrt {{{{K_2}} \over {{K_1}}}} $$
Explanation
$$ \because $$ $${V_{\max }} = A\omega $$
Given, $${\omega _1}{A_1} = {\omega _2}{A_2}$$
We know that $$\omega = \sqrt {{K \over m}} $$
$$ \therefore $$ $$\sqrt {{{{k_1}} \over m}} {A_1} = \sqrt {{{{k_2}} \over m}} {A_2}$$
$$ \Rightarrow $$ $${{{A_1}} \over {{A_2}}} = \sqrt {{{{k_2}} \over {{k_1}}}} $$
Given, $${\omega _1}{A_1} = {\omega _2}{A_2}$$
We know that $$\omega = \sqrt {{K \over m}} $$
$$ \therefore $$ $$\sqrt {{{{k_1}} \over m}} {A_1} = \sqrt {{{{k_2}} \over m}} {A_2}$$
$$ \Rightarrow $$ $${{{A_1}} \over {{A_2}}} = \sqrt {{{{k_2}} \over {{k_1}}}} $$
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