Two particles A and B of equal masses are suspended from two massless springs of spring constants

Question:

Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $\mathrm{K}_{1}$ and $\mathrm{K}_{2}$ respectively.If the maximum velocities during oscillations are equal, the ratio of the amplitude of $\mathrm{A}$ and $\mathrm{B}$ is

  1. (1) $\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}$

  2. (2) $\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}$

  3. (3) $\sqrt{\frac{K_{1}}{K_{2}}}$

  4. (4) $\sqrt{\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}}$


Correct Option: , 4

Solution:

(4)

$\mathrm{A}_{1} \omega_{1}=\mathrm{A}_{2} \omega_{2}$

$\mathrm{A}_{1} \sqrt{\frac{\mathrm{k}_{1}}{\mathrm{~m}}}=\mathrm{A}_{2} \sqrt{\frac{\mathrm{k}_{2}}{\mathrm{~m}}}$

$\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\sqrt{\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}}$

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