If the radius of curvature of the path of two particles of same...

Given ${m}_{1} = {m}_{2}$ and $\dfrac{{r}_{1}}{{r}_{2}} = \dfrac{3}{4}$

As centripetal force is given by $F = \dfrac{m{v}^{2}}{r}$, in order to have the same centripetal force, we can write

${F}_{1} = {F}_{2} ⇒ \dfrac{{m}_{1}{v}_{1}^{2}}{{r}_{1}} = \dfrac{{m}_{2}{v}_{2}^{2}}{{r}_{2}} ⇒ \dfrac{{v}_{1}}{{v}_{2}} = \sqrt{\dfrac{{r}_{1}}{{r}_{2}}} = \dfrac{\sqrt{3}}{2}$