Two bodies A and B , having the same mass, are hung from separate...

The maximum velocity of a body 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, the angular frequency is $\omega = \sqrt{\dfrac{k}{m}}$.

Given that the maximum velocities of both bodies are identical, $v_{max, A} = v_{max, B}$.

Therefore, $A_A \omega_A = A_B \omega_B$.

Substituting the expressions for $\omega$, we get $A_A \sqrt{\dfrac{k_1}{m}} = A_B \sqrt{\dfrac{k_2}{m}}$.

Since the masses are the same, this simplifies to $A_A \sqrt{k_1} = A_B \sqrt{k_2}$.

Rearranging to find the ratio of the amplitudes, we obtain $\dfrac{A_A}{A_B} = \sqrt{\dfrac{k_2}{k_1}}$.