Two particles A and B of masses ′m′ and ′2m′ are suspended from two massless springs of force constants K1 and K2. During their oscillation, if their maximum velocities are equal, then ratio of amplitudes of A and B is :
A
√K1K2
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B
√K22K1
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C
√K2K1
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D
√2K1K2
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Solution
The correct option is A√K22K1 Since by energy conservation,
12KA2=12Mv2max.
Hence, K1A21K2A22=m2m=12, since maximum velocities are same in both cases.