1. Given displacement equation

   The displacement is $x = 4(\cos \pi t + \sin \pi t).$

2. Rewrite in standard SHM form

   For an expression of the form $a\cos \theta + b\sin \theta,$ the maximum value is $\sqrt{a^2+b^2}.$

   Here, $x = 4\cos \pi t + 4\sin \pi t.$

   So, $a=4, \quad b=4.$

3. Calculate amplitude

   Therefore amplitude is $A=\sqrt{4^2+4^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}.$

4. Alternative verification

   We can also write $\cos \pi t + \sin \pi t = \sqrt{2}\sin\left(\pi t + \frac{\pi}{4}\right),$ hence $x = 4\sqrt{2}\sin\left(\pi t + \frac{\pi}{4}\right).$

   Comparing with standard SHM form $x=A\sin(\omega t+\phi),$ we get $A=4\sqrt{2}.$

5. Check options

   • A: $-4$ → amplitude cannot be negative, incorrect
   • B: $4$ → incorrect
   • C: $4\sqrt{2}$ → correct
   • D: $8$ → incorrect

Final Answer: Option C, $4\sqrt{2}$