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(B) The maximum velocity of a particle in simple harmonic motion is given by $V_{\max} = A \omega$, where $A$ is the amplitude and $\omega = \sqrt{\fr

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$\sqrt{\frac{K_2}{2 K_1}}$

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(B) The maximum velocity of a particle in simple harmonic motion is given by $V_{\max} = A \omega$, where $A$ is the amplitude and $\omega = \sqrt{\frac{K}{m}}$ is the angular frequency.For particle $A$: $(V_{\max})_A = A_A \sqrt{\frac{K_1}{m}}$.For particle $B$: $(V_{\max})_B = A_B \sqrt{\frac{K_2}{2m}}$.Given that $(V_{\max})_A = (V_{\max})_B$, we have:$A_A \sqrt{\frac{K_1}{m}} = A_B \sqrt{\frac{K_2}{2m}}$.Squaring both sides:$A_A^2 \frac{K_1}{m} = A_B^2 \frac{K_2}{2m}$.Rearranging for the ratio of amplitudes $\frac{A_A}{A_B}$:$\frac{A_A^2}{A_B^2} = \frac{K_2}{2m} \cdot \frac{m}{K_1} = \frac{K_2}{2K_1}$.Therefore, $\frac{A_A}{A_B} = \sqrt{\frac{K_2}{2K_1}}$.

Two particles $A$ and $B$ of masses $m$ and $2m$ are suspended from massless springs of force constants $K_1$ and $K_2$. During their oscillation, if their maximum velocities are equal, then the ratio of amplitudes of $A$ and $B$ is

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