Physics Question from JEE Main 2025 · 2 April, Shift 1

A particle is subjected to two simple harmonic motions as: $x_1 = \sqrt{7} \sin 5t \, \text{cm}$ and $x_2 = 2 \sqrt{7} \sin \left( 5t + \frac{\pi}{3} \right) \, \text{cm}$ where $x$ is displacement and $t$ is time in seconds. The maximum acceleration of the particle is $x \times 10^{-2} \, \text{m/s}^2$. The value of $x$ is: