Two objects A and B of equal mass are suspend | vedclass

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(B) The maximum kinetic energy $(KE_{max})$ of an object undergoing simple harmonic motion is given by the formula $KE_{max} = \frac{1}{2} k A^2$,wher

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$\sqrt{\frac{k_B}{k_A}}$

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(B) The maximum kinetic energy $(KE_{max})$ of an object undergoing simple harmonic motion is given by the formula $KE_{max} = \frac{1}{2} k A^2$,where $k$ is the spring constant and $A$ is the amplitude.Given that the maximum kinetic energies of objects $A$ and $B$ are equal,we have:$\frac{1}{2} k_A A_A^2 = \frac{1}{2} k_B A_B^2$Rearranging the terms to find the ratio of amplitudes $\frac{A_A}{A_B}$:$\frac{A_A^2}{A_B^2} = \frac{k_B}{k_A}$Taking the square root on both sides,we get:$\frac{A_A}{A_B} = \sqrt{\frac{k_B}{k_A}}$

Two objects $A$ and $B$ of equal mass are suspended from two springs of spring constants $k_A$ and $k_B$. If the objects oscillate vertically in such a manner that their maximum kinetic energies are equal,then the ratio of amplitudes of $A$ and $B$ is

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