Riding The Sinusoid

Because there is no friction we can write the conservation of energy and find the velocity of the body at any point given as follows: $mg{{y}_{0}}=\frac{1}{2}m{{v}^{2}}+mgy(x)$ $v=\sqrt{2g\left( {{y}_{0}}-y(x) \right)}$ But $v=\frac{ds}{dt}$ where $ds$ is the arc element of the curve. In our case: $ds=\sqrt{1+{{{{y}'}}^{2}}(x)}dx$ Substituting in the formula for $v$ we get: $\frac{dx}{dt}\sqrt{1+{{{{y}'}}^{2}}(x)}=\sqrt{2g\left( {{y}_{0}}-y(x) \right)}$ Rearranging: $dt=\frac{\sqrt{1+{{{{y}'}}^{2}}(x)}}{\sqrt{2g\left( {{y}_{0}}-y(x) \right)}}dx$ Now we will integrate only on a half of the motion and then we will multiply our result by 2: $\int\limits_{0}^{T}{dt}=\int\limits_{0.7}^{2\pi -0.7}{\frac{\sqrt{1+{{\sin }^{2}}(x)}}{\sqrt{20\left( \cos (0.7)-\cos (x) \right)}}dx}$ $T=2.00184$ So the final result is $P=2T=4.0036s$