The speed of an object in a mass-spring system is given under the function
[tex]v = \pm \sqrt{\frac{k}{m}(A^2-x^2)}[/tex]
Here,
m = mass
k = Spring constant
A = Amplitude
x = Position
When the position is at the equilibrium point (x = 0), the speed will be maximum, and could even be expressed as
[tex]v_{max}= A\sqrt{\frac{x}{m}}[/tex]
So the correct answer is B.