1. The given periodic motion is $y = \sin \omega t + \cos \omega t$

2. To find the amplitude, rewrite the expression in the standard single sine/cosine form: $a\sin \omega t + b\cos \omega t = R\sin(\omega t + \phi)$ where $R = \sqrt{a^2+b^2}$

3. Here, $a=1,\qquad b=1$ so $R = \sqrt{1^2+1^2} = \sqrt{2}$

4. Therefore, the motion can be written as $y = \sqrt{2}\sin\left(\omega t + \frac{\pi}{4}\right)$ Hence the amplitude is $\boxed{\sqrt{2}}$

5. Checking options:

   • A: $1$ ❌
   • B: $\sqrt2$ ✅
   • C: $0.5$ ❌
   • D: $2$ ❌

So the correct option is B.