1. Given SHMs

   $x_1=\sqrt{7}\sin 5t\;\text{cm}$ $x_2=2\sqrt{7}\sin\left(5t+\frac{\pi}{3}\right)\;\text{cm}$

   Both motions have the same angular frequency:

   $\omega=5\ \text{rad s}^{-1}$

2. Resultant amplitude

   For two SHMs of same frequency and phase difference $\phi=\frac{\pi}{3}$, resultant amplitude is

   $A=\sqrt{A_1^2+A_2^2+2A_1A_2\cos\phi}$

   where

   $A_1=\sqrt{7}\ \text{cm}, \qquad A_2=2\sqrt{7}\ \text{cm}$

   So,

   $A=\sqrt{(\sqrt{7})^2+(2\sqrt{7})^2+2(\sqrt{7})(2\sqrt{7})\cos\frac{\pi}{3}}$

   $A=\sqrt{7+28+2\cdot 14\cdot \frac{1}{2}}$

   $A=\sqrt{7+28+14}$

   $A=\sqrt{49}=7\ \text{cm}$

   Hence resultant motion has amplitude

   $A=7\ \text{cm}=7\times 10^{-2}\ \text{m}$

3. Maximum acceleration in SHM

   For SHM,

   $a=-\omega^2 x$

   so maximum acceleration is

   $a_{\max}=\omega^2 A$

   Therefore,

   $a_{\max}=5^2\times 7\times 10^{-2}$

   $a_{\max}=25\times 7\times 10^{-2}$

   $a_{\max}=175\times 10^{-2}\ \text{m s}^{-2}$

4. Compare with the required form

   Given maximum acceleration is $x\times 10^{-2}\ \text{m s}^{-2}$.

   Hence,

   $x=175$

5. Option check

   • A: $5\sqrt{7}$ ❌
   • B: $125$ ❌
   • C: $25\sqrt{7}$ ❌
   • D: $175$ ✅

Therefore, the correct answer is Option D.