1. For a mass-spring system executing SHM, the maximum speed is $v_{\max}=A\omega$ where $A$ is amplitude and $\omega$ is angular frequency.

2. For a vertical spring-mass system of mass $m$ and spring constant $k$, the angular frequency is $\omega=\sqrt{\frac{k}{m}}$ Note: vertical orientation only shifts the equilibrium position; it does not change the frequency.

3. Since the two bodies have equal masses and equal amplitudes, let the common amplitude be $A$.

   For body A: $v_{A,\max}=A\sqrt{\frac{k_1}{m}}$

   For body B: $v_{B,\max}=A\sqrt{\frac{k_2}{m}}$

4. Therefore, the required ratio is \frac{v_{A,\max}}{v_{B,\max}}=rac{A\sqrt{k_1/m}}{A\sqrt{k_2/m}}=\sqrt{\frac{k_1}{k_2}}

5. Now check the options:

   • A: $\sqrt{\frac{k_2}{k_1}}$ — incorrect
   • B: $\sqrt{\frac{k_1}{k_2}}$ — correct
   • C: $\frac{k_2}{k_1}$ — incorrect
   • D: $\frac{k_1}{k_2}$ — incorrect

Hence, the correct answer is Option B.