Determine Parameter x from Simple Harmonic Motion Time Period

The velocity $v$ of a particle executing simple harmonic motion (SHM) as a function of its displacement $x$ is given by the standard relation: $v^2 = \omega^2(A^2 - x^2)$ $v^2 = \omega^2 A^2 - \omega^2 x^2$

where:

• $\omega$ is the angular frequency of the motion,
• $A$ is the amplitude of the motion.

Comparing the given equation $v^2 = 50 - x^2$ with the standard equation: $\omega^2 = 1 \implies \omega = 1 \text{ rad/s}$

The time period $T$ of a simple harmonic motion is given by: $T = \frac{2\pi}{\omega}$

Substituting $\omega = 1 \text{ rad/s}$: $T = \frac{2\pi}{1} = 2\pi \text{ s}$

Taking $\pi \approx \frac{22}{7}$: $T = 2 \times \frac{22}{7} = \frac{44}{7} \text{ s}$

According to the question, the time period of motion is given as $\frac{x}{7} \text{ s}$. Equating the two expressions: $\frac{x}{7} = \frac{44}{7}$

$x = 44$