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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 $\omega$ is the angul

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

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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 $\omega$ is the angular frequency.For a spring-mass system,$\omega = \sqrt{\frac{K}{m}}$.Thus,$V_{\max} = A\sqrt{\frac{K}{m}}$.Given for particle $A$: mass $m_A = m$,force constant $K_A = K_1$,amplitude $A_A$.Given for particle $B$: mass $m_B = 2m$,force constant $K_B = K_2$,amplitude $A_B$.Since $(V_{\max})_A = (V_{\max})_B$,we have:$A_A \sqrt{\frac{K_1}{m}} = A_B \sqrt{\frac{K_2}{2m}}$$\frac{A_A}{A_B} = \sqrt{\frac{K_2}{2m} \cdot \frac{m}{K_1}}$$\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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