A block of mass m is connected to two identical massless springs,...

The maximum speed of a block in simple harmonic motion is given by v_\max = A\omega = A\sqrt{\dfrac{k_{\text{eq}}}{m}}.

In the first case, the springs are in series. The equivalent spring constant is $k_{\text{series}} = \dfrac{k \times k}{k + k} = \dfrac{k}{2}$.

In the second case, the springs are in parallel. The equivalent spring constant is $k_{\text{parallel}} = k + k = 2k$.

Since the amplitude $A$ and mass $m$ are the same in both cases, the maximum speed is directly proportional to $\sqrt{k_{\text{eq}}}$.

Therefore, the ratio of the maximum speeds is:
$\dfrac{v_{\text{series}}}{v_{\text{parallel}}} = \sqrt{\dfrac{k_{\text{series}}}{k_{\text{parallel}}}} = \sqrt{\dfrac{k/2}{2k}} = \sqrt{\dfrac{1}{4}} = \dfrac{1}{2}$.

The ratio is $1 : 2$.

Answer: $1 : 2$