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

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$\sqrt{K_A} : \sqrt{K_B}$

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(B) The maximum velocity $v_{max}$ 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, $\omega = \sqrt{\frac{K}{m}}$.Given that the masses are equal $(m_A = m_B = m)$, the angular frequencies are $\omega_A = \sqrt{\frac{K_A}{m}}$ and $\omega_B = \sqrt{\frac{K_B}{m}}$.Since the maximum velocities are equal, we have $A_A \omega_A = A_B \omega_B$.Substituting the expressions for $\omega$, we get $A_A \sqrt{\frac{K_A}{m}} = A_B \sqrt{\frac{K_B}{m}}$.Simplifying this, $A_A \sqrt{K_A} = A_B \sqrt{K_B}$.The ratio of the amplitude of '$B$' to that of '$A$' is $\frac{A_B}{A_A} = \frac{\sqrt{K_A}}{\sqrt{K_B}} = \sqrt{\frac{K_A}{K_B}}$.

Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $K_A$ and $K_B$ respectively. The maximum velocities of the particles during oscillations are equal. The ratio of the amplitude of '$B$' to that of '$A$' is

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