JEE MAIN - Physics (2022 - 24th June Evening Shift - No. 5)
Two massless springs with spring constants 2 k and 9 k, carry 50 g and 100 g masses at their free ends. These two masses oscillate vertically such that their maximum velocities are equal. Then, the ratio of their respective amplitudes will be :
1 : 2
3 : 2
3 : 1
2 : 3
Explanation
$${\omega _1}{A_1} = {\omega _2}{A_2}$$
$$ \Rightarrow {{{A_1}} \over {{A_2}}} = {{{\omega _2}} \over {{\omega _1}}}$$
$$ = \sqrt {{{{k_2}} \over {{m_2}}}} \times \sqrt {{{{m_1}} \over {{k_1}}}} = \sqrt {{{9k} \over {100}} \times {{50} \over {2k}}} = {3 \over 2}$$
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