To find the value of $$x$$ that represents the time period of oscillation in this simple harmonic motion (SHM) scenario, we first recall the general formula for the time period ($T$) of a mass-spring system undergoing SHM, which is given by:

$$T = 2\pi \sqrt{\frac{m}{k}}$$

Here,

$m$ is the mass of the particle, which is $$0.50 \, \mathrm{kg}$$ in this case,

$k$ is the force constant of the spring or the spring constant, which is given as $$50 \, \mathrm{Nm^{-1}}$$,

and $T$ represents the time period of oscillation.

Given in the problem, $$T = \frac{x}{35} \, \mathrm{s}$$ and we are provided with the approximation $$\pi = \frac{22}{7}$$.

Substituting the given values into the formula for $$T$$:

$$\frac{x}{35} = 2 \times \frac{22}{7} \times \sqrt{\frac{0.50}{50}}$$

To simplify this, we first calculate the square root:

$$\sqrt{\frac{0.50}{50}} = \sqrt{\frac{1}{100}} = \frac{1}{10}$$

Substituting back, we get:

$$\frac{x}{35} = 2 \times \frac{22}{7} \times \frac{1}{10}$$

Multiplying the terms on the right side:

$$\frac{x}{35} = \frac{44}{70}$$

$$\frac{x}{35} = \frac{22}{35}$$

Multiplying both sides by $$35$$ to solve for $$x$$:

$$x = 22$$

Therefore, the value of $$x$$ that represents the time period of oscillation is $$22$$ seconds.