This problem is fairly convenient to do with Lagrangian mechanics. Let $x$ be the horizontal coordinate of the particle on the right.

$x = x \\ y = x^{-1} \\ \dot{x} = \dot{x} \\ \dot{y} = -x^{-2} \dot{x}$

Kinetic energy:

$T = 2 \frac{1}{2} m v^2 = 2 \frac{1}{2} m (\dot{x}^2 + \dot{y}^2) = m(1 + x^{-4}) \dot{x}^2$

Potential energy:

$V = 2 m g x^{-1} + \frac{1}{2} k (2x - \ell)^2$

Lagrangian:

$L = T - V = m(1 + x^{-4}) \dot{x}^2 - 2 m g x^{-1} - \frac{1}{2} k (2x - \ell)^2$

Equation of motion:

$\frac{d}{dt} \frac{\partial{L}}{\partial{\dot{x}}} = \frac{\partial{L}}{\partial{x}}$

Evaluating results in:

$\ddot{x} = \frac{4 m \dot{x}^2 x^{-5} + 2 m g x^{-2} - 2 k (2x - \ell)}{2m(1 + x^{-4}) }$

Numerically integrate until $\dot{x} < 0$ for the first time. This time corresponds to a half period. Simulation code is attached.

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import math

dt = 10.0**(-6.0)

m = 1.0
g = 10.0
L0 = 2.0*m
k = 2.0*m*g

###################################

t = 0.0

x = 0.5
xd = 0.0
xdd = 0.0

while xd >= 0.0:

    x = x + xd*dt
    xd = xd + xdd*dt

    right = 4.0*m*(xd**2.0)*(x**-5.0)
    right = right + 2.0*m*g*(x**-2.0)
    right = right - 2.0*k*(2.0*x - L0)

    left = 2.0*m*(1.0 + x**(-4.0))

    xdd = right/left


    t = t + dt

###################################

T = 2.0*t

print dt
print (20.0*T)

#>>> 
#1e-05
#27.3884
#>>> ================================ RESTART ================================
#>>> 
#1e-06
#27.38756
#>>> 

Finding $h$ when it falls

Applying Conservation of Energy

$\frac12k(1)^2 + 2mgh = \frac12k\bigg(\frac{2}{2-h}-2\bigg)^2$ $\frac122mg + 2mgh = 4mg\bigg(\frac{h-1}{h-2}\bigg)^2$ $2h + 1 = 4\bigg(\frac{h-1}{h-2}\bigg)^2$ $\boxed{h = 1.5 m}$

Finding $ds$

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

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

$ds = \frac{\sqrt{y^4 + 1}}{y^2}dy$

Finding Time period $T$

Applying Conservation of Energy when they have fallen through height $a$

$\frac12k(1)^2 + 2mga = 2\frac12mv^2 + \frac12k\bigg(\frac{2}{2-a}-2\bigg)^2$

$mg + 2mga = mv^2 + 4mg\bigg(\frac{1-a}{2-a}\bigg)^2$

$v = \sqrt{g}×\frac{\sqrt{a(2a-3)(a-4)}}{a-2}$

Putting $a = 2-y$

$v = \sqrt{g}×\frac{\sqrt{(2-y)(1-2y)(-2-y)}}{-y}$

$\frac{ds}{dt} = \sqrt{g}×\frac{\sqrt{(y-2)(2y-1)(y+2)}i}{-y}$

$\frac{ds}{\sqrt{g}×\frac{\sqrt{(y-2)(2y-1)(y+2)}i}{-y}} = dt$

$\frac{\frac{\sqrt{y^4 + 1}}{y^2}dy}{\sqrt{g}×\frac{\sqrt{(y-2)(2y-1)(y+2)}i}{-y}} = dt$

$t = \frac{1}{\sqrt{g}}\int_2^{2-h}-\sqrt{\frac{y^4 + 1}{(y-2)(2y-1)(y+2)}} × \frac{1}{yi} dy$

$t = \frac{1}{\sqrt{10}} \int_2^{0.5}-\sqrt{\frac{y^4 + 1}{(y-2)(2y-1)(y+2)}} × \frac{1}{yi} dy$

$\boxed{t = 0.6846 sec}$

$\text{Time Period } T = 2t$

$\color{#3D99F6}{\boxed{T \approx 1.36937 sec}}$

Therefore, $\text{answer } = 20T = 27.387 sec$

A slightly different approach. Let the coordinates of the particle on the right be:

$(x_1,y_1) = \left(s,\frac{1}{s}\right)$

Let the coordinates of the particle on the left be:

$(x_2,y_2) = \left(-s,\frac{1}{s}\right)$

Total kinematic energy of the system is:

$\mathcal{T}=\frac{m}{2}\left(\dot{x}_1^2 + \dot{y}_1^2\right) + \frac{m}{2}\left(\dot{x}_2^2 + \dot{y}_2^2\right)$

The total potential energy of the system is the sum of spring PE and gravitational PE. This comes out to be:

$\mathcal{V} = mgy_1 + mgy_2 + \frac{K}{2}\left(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} - L\right)^2$

Plugging in the parameterization and simplifying expressions, one gets:

$\mathcal{T} = m\left(\frac{s^4+1}{s^4}\right)\dot{s}^2$

$\mathcal{V} = 4mg(s-1)^2 + \frac{2mg}{s}$

The energy conservation principle is then applied as such:

$\mathcal{T}_{initial} + \mathcal{V}_{initial} = \mathcal{T} + \mathcal{V}$

Simplifying the above expression gives:

$\dot{s} = \sqrt{\frac{1+s^4}{10s^3(5s-2)-40s^4(s-1)^2}}$

Now, this equation can be separated and integrated. But before doing so, the bounds of oscillation are required to be found. It can be seen that the initial total PE is at $s=0.5$ and the same value is obtained again when $s=2$ . This can be observed by inspection or by the plotting the potential energy as a function of $s$ . So the system oscillated between $0.5 \le s \le 2$ . The time period can be found by separating the variables and integrating between the bounds. The resulting expression must be multiplied by 2 as movement from one bound to another occurs in half a time period.

Thus:

$T = 2\int_{0.5}^{2}\sqrt{\frac{10s^3(5s-2)-40s^4(s-1)^2}{1+s^4}} \ ds$

The rest is all numerics. The answer comes out to be: $\boxed{20T \approx 27.38743362322671}$