Step I : Determine the radius of curvature of given curve at point P (x, y).

∵       $y=a\text{sin}\frac{x}{\alpha }\therefore \frac{dy}{dx}=\frac{a}{\alpha }\text{cos}\frac{x}{\alpha }$

∴      $\frac{d^{2}y}{d{x}^2}=-\frac{a}{{\alpha }^2}\text{sin}\frac{x}{\alpha }$

∴    the radius of curvature is  $R=\frac{{\left[1+{\left(\frac{dy}{dx}\right)}^2\right]}^{3/2}}{\frac{d^{2}y}{d{x}^2}}$

                                             $R=\frac{{\left[1+\frac{a^2}{{\alpha }^2}{\text{cos}}^2\frac{x}{a}\right]}^{3/2}}{-\frac{a}{{\alpha }^2}\text{sin}\frac{x}{\alpha }}$        ...(i)

Step II: Relate the friction and centripetal acceleration. In this case, centripetal acceleration is provided by force of friction.

∵                              $f \le f_L$     or$f \le kmg$

or                              $\frac{{\text{mv}}^2}{R}\le \text{kmg}\text{or}\frac{v^2}{R}\le \text{kg}$

∴                              $\frac{v_{\text{max}}^2}{R_{\text{min}}}=k\text{g}$

∴                              $v_{\text{max}}=\sqrt{k\text{g}{R}_{\text{min}}}$

From equation (i), for the minimum value of the radius of curvature,

                              $\frac{x}{\alpha }=\frac{\pi }{2}$

∴                              $R_{\text{min}}=\frac{{\alpha }^2}{a}$                             (from equation. (i))

∴                              $v_{\text{max}}=\sqrt{\frac{k\text{g}{\alpha }^2}{a}}=\alpha \sqrt{\frac{k\text{g}}{a}}$