Two particles A and B of equal masses are suspended from two mass less spring of spring constants K1 and K2 respectively. If the maximum velocities, during oscillations are equal. The ratio of amplitudes of 'A' and 'B'.

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K1K2

K2K1

answer is C.

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Detailed Solution

vmax=aw =a2πT =a2π2πmk vmax =akm Hence vmax1vmax2=a1a2K1/mK2/m given vmax1=vmax2 a1a2=K2K1

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