The velocity of a particle executing simple harmonic motion along x...

The given equation is $v^2 = 50 - x^2$

Differentiating with respect to $x$:

$2v \dfrac{dv}{dx} = -2x$

$v \dfrac{dv}{dx} = -x$

Since acceleration $a = v \dfrac{dv}{dx}$, we get:

$a = -x$

Comparing this with the standard equation of SHM $a = -\omega^2 x$, we get:

$\omega^2 = 1 \Rightarrow \omega = 1$ rad/s

The time period of the motion is given by:

$T = \dfrac{2\pi}{\omega} = 2\pi$

Taking $\pi = \dfrac{22}{7}$, we get:

$T = 2 \times \dfrac{22}{7} = \dfrac{44}{7}$ s

Given that the time period is $\dfrac{x}{7}$ s, comparing the two gives:

$x = 44$

Answer: $44$