JEE MAIN - Physics (2003 - No. 41)
Two particles $$A$$ and $$B$$ of equal masses are suspended from two massless springs of spring 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
$$\sqrt {{{{k_1}} \over {{k_2}}}} $$
$${{{{k_2}} \over {{k_1}}}}$$
$$\sqrt {{{{k_2}} \over {{k_1}}}} $$
$${{{{k_1}} \over {{k_2}}}}$$
解释
Maximum velocity during $$SHM$$ $$ = A\omega = A\sqrt {{k \over m}} $$
$$\left[ {\,\,} \right.$$ $$\therefore$$ $$\omega = \sqrt {{k \over m}} $$ $$\left. {\,\,} \right]$$
Here the maximum velocity is same and $$m$$ is also same
$$\therefore$$ $${A_1}\sqrt {{k_1}} = {A_2}\sqrt {{k_2}} $$
$$\therefore$$ $${{{A_1}} \over {{A_2}}} = \sqrt {{{{k_2}} \over {{k_1}}}} $$
$$\left[ {\,\,} \right.$$ $$\therefore$$ $$\omega = \sqrt {{k \over m}} $$ $$\left. {\,\,} \right]$$
Here the maximum velocity is same and $$m$$ is also same
$$\therefore$$ $${A_1}\sqrt {{k_1}} = {A_2}\sqrt {{k_2}} $$
$$\therefore$$ $${{{A_1}} \over {{A_2}}} = \sqrt {{{{k_2}} \over {{k_1}}}} $$