Finding compression in spring

When ring fall height $h$ , it's coordinates become $(-2\sqrt{1 - h}, 1 - h)$ . Let it's compression be $x$

$x = 2 - \sqrt{(0- (-2\sqrt{1 - h}))^2 + (1- ( 1 - h))^2}$ $\boxed{x = h}$

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Finding $ds$

Equation of wedge $y = \frac{x^2}{4}$ or $x = 2\sqrt{y}$ and $\displaystyle \frac{dx}{dy} = \frac{1}{\sqrt{y}}$

$ds = \sqrt{1 + \bigg(\frac{dx}{dy}\bigg)^2}dy$

$ds = \sqrt{1 + \frac{1}{y}}dy$

Now, when $y = 1 - h$ , $dy = -dh$

$ds = \sqrt{1 + \frac{1}{1 - h}}dh$

$\boxed{ds = \sqrt{\frac{2 - h}{1 - h}}dh}$

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Finding time

By conservation of energy

$mgh = \frac12mv^2 + \frac12kx^2$

Put $x = h$ and $k = mg$

$mgh = \frac12mv^2 + \frac12mgh^2$

$v = \sqrt{2gh - gh^2}$

$\frac{ds}{dt} = \sqrt{gh(2 - h)}$

$dt = \frac{ds}{\sqrt{gh(2 - h)}}$

$dt = \frac{\sqrt{\frac{2 - h}{1 - h}}}{\sqrt{gh(2 - h)}}dh$

$t = \frac{1}{\sqrt{g}}\int_0^1\sqrt{\frac{1}{h(1-h)}}dh$

$t = \frac{2}{\sqrt{g}}\bigg[sin^{-1}(\sqrt{h})\bigg]_0^1$

$t = \frac{1}{\sqrt{g}}\pi$

Putting $g = 10m/s^2$ , we get

$t = \frac{1}{\sqrt{10}}\pi$

$\text{Time period } T = 4t$

$T = \frac{4}{\sqrt{10}}\pi$

$\color{#3D99F6}{\boxed{T = \sqrt{\frac85}\pi}}$

Therefore $a = 8, b = 5 \text{ and } a + b = 13$

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Visualisation: Time Period $T\approx4.03 \text{ sec}$

The position vector and velocity of the particle are $\mathbf{r} \; = \; \binom{x}{\frac14x^2} \hspace{2cm} \dot{\mathbf{r}} \; = \; \binom{1}{\frac12x}\dot{x}$ so the kinetic and gravitational potential energies of the system are $T \; = \; \tfrac12m\big(1 + \tfrac14x^2\big)\dot{x}^2 \hspace{2cm} V \; = \; \tfrac14mgx^2$ In general, the length of the spring is $L \; = \; \sqrt{x^2 + (1 - \tfrac14x^2)^2} \; = \; 1 + \tfrac14x^2$ (this reflects the fact that $(0,1)$ is the focus, and the line $y=-1$ the directrix, of the parabola) and so the elastic potential energy of the system is $E \; = \; \tfrac12k(L-2)^2 \; = \; \tfrac12k\big(1 - \tfrac14x^2\big)^2$ Conservation of energy tells us that $T + V + E = mg$ , so that $\begin{aligned}\tfrac12m\big(1 + \tfrac14x^2\big)\dot{x}^2 & = \; mg - V - E \; = \; \tfrac{1}{32}\big(4 - x^2)(8mg - 4k + kx^2)\big) \\ \big(1 + \tfrac14x^2\big)\dot{x}^2 & = \; g\big(1 - \tfrac14x^2\big)\big(1 + \tfrac14x^2\big) \\ \dot{x}^2 & = \; \tfrac52(4 - x^2) \end{aligned}$ putting $k = mg$ and $g=10$ . The the particle oscillates between $x=2$ and $x=-2$ with period $T \; = \; 2\int_{-2}^2 \sqrt{\frac{2}{5(4-x^2)}}\,dx \; = \; \pi\sqrt{\tfrac85}$ making the answer $8+5 = \boxed{13}$ .