# Descent on an inclined wavy frictionless track [closed]

The classical Brachistochrone was actually counterintuitive wherein the time of descent is lesser (the least) for the cycloid than that of the corresponding straight inclined track.

Let an inclined and wavy track be given by $$y= 1-xe^{\epsilon \cos^2(7x\pi/2)}~~~(1)$$ for $$\epsilon=0$$ this is a simple inclined, straight track and the time taken by for a particle to fall down is $$2 \tau$$, where $$\tau=\frac{1}{\sqrt{g}}$$ in CGS.

The question is on which track the time of descent will be lesser: on the red ($$\epsilon=+0.06$$) or on the blue ($$\epsilon=-0.06$$) track? Justify your answer with or without a calculation. See the figure below for the two complementary inclined tracks.

• I've removed a discussion that should have taken place on Physics Meta.
– rob
Oct 14, 2020 at 19:01

From energy conservation: $$\frac{1}{2}m(\dot{x}^2 + \dot{y}^2)+mgf(x) = \mathrm{const}$$ if you start at rest at $$x=0$$, the constant is $$mgf(0)$$. With some drawing you can convince yourself that $$\dot{y} = f'(x)\dot{x}$$. Then, we get: $$\left(\frac{dx}{dt}\right)^2 = \frac{2(f(0)-f(x))g}{1+f'(x)^2}$$ which gives: $$T = \int_0^1dx\left(\frac{1+f'(x)^2}{2(f(0)-f(x))g}\right)^{1/2}$$ then you can do numerical integration or whatever you want. It would be very surprising to have $$t_R > t_B$$ though, since at any $$x$$ the velocity of the mobile on the blue curve is be higher than the velocity of the mobile on the red curve by energy conservation, and with the naked eye it does not look like the two have different lengths
• $f'(x)$ needs to be squared. Finally, $t_R< t_B$, you may see my edit of the answer. Oct 15, 2020 at 10:51
The time of descent over a vertical curve given by $$y=f(x)$$ is found as $$t=\int_{x_1}^{x_2}\sqrt{\frac{1+f'(x)^2}{2g(f(x_1)-f(x))}} dx.~~~~(1)$$ Here $$f(x)=1-xe^{\epsilon \cos^2(7x \pi/2)}, x \in [0,1]~~~~(2)$$
This integral can be carried out numerically to get $$t_R=1.9951 \tau$$, for the red track $$(\epsilon=+0.06)$$. For the blue track it comes to a lesser value of $$t_B=2.0280 \tau$$. Here $$\tau=\frac{1}{\sqrt{g}}$$ in CGS. For the linear track $$y=1-x$$, $$t_L= 2 \tau$$. The time of free-fall from a distance $$\sqrt{2}$$ under the gravity is $$t_F=\sqrt{2\sqrt{2}}\tau=1.6817 \tau$$.
Result that $$t_R demonstrates a Brachistochrone effect. But the result $$t_B>t_L$$ requites an explanation. In fact if enlarged the blue tack is convex initially whereas the red one is concave.