The correct answer is $\boxed{\text{B. }1.57\text{ m/s}}$.
The maximum velocity of a particle in simple harmonic motion is reached when the particle is at its extreme position. In this case, the particle is 0.5 m from the equilibrium position, and its velocity is given by:
$$v = \omega\sqrt{A^2 – x^2}$$
where $\omega$ is the angular frequency of the motion, $A$ is the amplitude of the motion, and $x$ is the displacement of the particle from the equilibrium position.
We are given that the time period of the motion is 1 s, so the angular frequency is:
$$\omega = \frac{2\pi}{T} = \frac{2\pi}{1\text{ s}} = 2\pi\text{ rad/s}$$
We are also given that the amplitude of the motion is 1 m, so the maximum velocity of the particle is:
$$v = \omega\sqrt{A^2 – x^2} = 2\pi\sqrt{1^2 – 0.5^2} = 2\pi\sqrt{0.75} = 1.57\text{ m/s}$$
The other options are incorrect because they do not represent the maximum velocity of the particle in simple harmonic motion.