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

The net displacement of the particle is the algebraic sum of the two given SHMs that have the same angular frequency $$\omega = 5 \text{ s}^{-1}$$:
$$x = x_1 + x_2 = \sqrt{7}\,\sin 5t + 2\sqrt{7}\,\sin\left(5t + \frac{\pi}{3}\right) \text{ cm}$$

When two SHMs with the same $$\omega$$ are added, the result is again an SHM with the same $$\omega$$ but with a new amplitude $$R$$ obtained from the phasor (vector) addition formula:
$$R = \sqrt{A_1^{2} + A_2^{2} + 2A_1A_2\cos\phi}$$
where
$$A_1 = \sqrt{7}\text{ cm},\quad A_2 = 2\sqrt{7}\text{ cm},\quad\phi = \frac{\pi}{3}$$

Compute each term:
$$A_1^{2} = 7,\quad A_2^{2} = 28,$$
$$2A_1A_2\cos\phi = 2(\sqrt{7})(2\sqrt{7})\cos\frac{\pi}{3} = 4\cdot7\cdot\frac12 = 14$$

Therefore,
$$R = \sqrt{7 + 28 + 14} = \sqrt{49} = 7\text{ cm}$$

The combined motion can thus be written as
$$x = 7\sin(5t + \delta)\text{ cm}$$
for some phase constant $$\delta$$.

The acceleration in SHM is given by $$a = -\omega^{2}x$$. Hence, the magnitude of the maximum acceleration is
$$a_{\max} = \omega^{2}R$$

Convert the amplitude to metres: $$R = 7\text{ cm} = 7\times10^{-2}\text{ m}$$.

Now substitute $$\omega = 5\text{ s}^{-1}$$:
$$a_{\max} = (5)^{2}\,(7\times10^{-2}) = 25 \times 0.07 = 1.75\text{ m s}^{-2}$$

The question states $$a_{\max} = x \times 10^{-2}\text{ m s}^{-2}$$. Write $$1.75$$ in that form:
$$1.75 = 175 \times 10^{-2}$$

Hence, $$x = 175$$.

Option A is correct.