Answer doesn't simplifying does!

Let's call the points with $8m/s, 7m/s, 4m/s$ as $A, B, C$ respectively.

Let $h$ be the displacement of point $A$ from the mean position.

It follows that the displacements of $B$ and $C$ from the mean position are $(h+1)$ and $(h+2)$ respectively.

Using: $v = \omega\sqrt{A^2 - x^2}$ where $v$ is instantaneous velocity, $A$ is amplitude from mean position, $\omega$ is the angular frequency and $x$ is displacement from mean position:

$8 = \omega\sqrt{A^2 - h^2} \implies 64 = A^2\omega^2 - h^2\omega^2 \hspace{2 cm} (1)$ $7 = \omega\sqrt{A^2 - (h+1)^2} \implies 49 = A^2\omega^2 - (h+1)^2\omega^2 \hspace{2 cm} (2)$ $4 = \omega\sqrt{A^2 - (h+2)^2} \implies 16 = A^2\omega^2 - (h+2)^2\omega^2 \hspace{2 cm} (3)$

Subtracting $(2)$ from $(1)$ :

$15 = (2h+1)\omega^2 \hspace{2 cm} (4)$

Subtracting $(3)$ from $(2)$ :

$33 = (2h+3)\omega^2 \hspace{2 cm} (5)$

Dividing $(5)$ by $(4)$ :

$\dfrac{33}{15} = \dfrac{2h+3}{2h+1}\implies h = \dfrac{1}{3}$

Substituting the above in $(4)$ :

$15 = \frac{15}{9}\omega^2 \implies \omega = 3$

Substituing the values of $h$ and $\omega$ in $(1)$ gives $A = \dfrac{\sqrt{65}}{3}$ .

The maximum velocity $V$ is at the mean position and is given by $V = A\omega$ .

$\therefore V = \sqrt{65} \implies \lfloor V \rfloor = \boxed{8}$