Spring Action

We note that the particle is subject to a Hooke's Law restoring force of magnitude $kx$ , and hence performs $SHM$ . Thus $x(t) \; =\; x_0 \cos \omega t \hspace{2cm} 0 \le t \le t_f \; =\; \tfrac{\pi}{2\omega}$ where $\omega^2 = \tfrac{k}{m}$ .

If $y(t)$ is any function such that $y(0)= y(t_f) = 0$ , then the Principle of Least Action tells us that the function $\begin{aligned} A(u) & = \; \int_0^{t_f} L_{x + uy}(t)\,dt \; = \; \int_0^{t_f}\left(\tfrac12m\big(\dot{x}(t) + u\dot{y}(t)\big)^2 - \tfrac12k\big(x(t) + uy(t)\big)^2\right)\,dt \\ & = \; \int_0^{t_f}\left(\tfrac12m\dot{x}(t)^2 - \tfrac12kx(t)^2\right)\,dt + u\int_0^{t_f}\big(m\dot{x}(t)\dot{y}(t) - kx(t)y(t)\big)\,dt + u^2\int_0^{t_f}\big(\tfrac12m\dot{y}(t)^2 - \tfrac12ky(t)^2\big)\,dt \\ & = \; \int_0^{t_f}L_x(t)\,dt + u\int_0^{t_f}\big(m\dot{x}(t)\dot{y}(t) - kx(t)y(t)\big)\,dt + u^2 \int_0^{t_f}L_y(t)\,dt \end{aligned}$ is minimized at $u=0$ , and hence that $\int_0^{t_f}\big(m\dot{x}(t)\dot{y}(t) - kx(t)y(t)\big)\,dt \; = \; 0$ In this case we also have $\int_0^{t_f}L_x(t)\,dt \; = \; -\int_0^{\frac{\pi}{2\omega}} \tfrac12mx_0^2\omega^2\cos2\omega t\,dt \; = \; 0$ and hence we have that $A(u) \; = \; u^2\int_0^{t_y}L_y(t)\,dt$ In this particular case we have $y(t) = t(t_f-t)$ , and hence $A(u) \; =\; \frac{mu^2\pi^3(40 - \pi^2)}{1920\omega^3}$ We are asked to evaluate $S + S_1 + S_2 + S_3 + S_4 \; =\; A(0) + A(2\beta) + A(\beta) + A(-\tfrac12\beta) + A(-\tfrac32\beta) \; = \; \frac{m\beta^2\pi^3(40 - \pi^2)}{256\omega^3}$ in the case $m=1$ , $\beta=1$ , $\omega = \sqrt{5}$ , and so we deduce that $S + S_1 + S_2 + S_3 + S_4 \; = \; \frac{\pi^3(40 - \pi^2)}{1280\sqrt{5}} = \boxed{0.3264070122...}$