The velocity of a particle executing simple harmonic motion along $$x$$-axis is described as $$v^2 = 50 - x^2$$, where $$x$$ represents displacement. If the time period of motion is $$\frac{x}{7}$$ s, the value of $$x$$ is _____.

For a particle in simple harmonic motion, the standard velocity-displacement relation is
$$v^2 = \omega^2\left(A^2 - x^2\right)$$
where $$A$$ is the amplitude and $$\omega$$ is the angular frequency.

The given relation is
$$v^2 = 50 - x^2$$.

Match coefficients with the standard form:

$$\omega^2 A^2 = 50 \quad\text{and}\quad \omega^2 = 1$$

From $$\omega^2 = 1$$ we get
$$\omega = 1\; \text{rad s}^{-1}$$.

The time period $$T$$ of SHM is
$$T = \frac{2\pi}{\omega} = 2\pi\; \text{s}$$.

The question states that the time period is also $$\dfrac{x}{7}\,\text{s}$$, so

$$\frac{x}{7} = 2\pi \;\Longrightarrow\; x = 14\pi.$$

Numerically, $$14\pi \approx 43.98 \approx 44.$$

Therefore, the required value of $$x$$ is 44.