Periodic Motion & SHM Basics question from MHT CET 2026 (15 Apr, Shift 2)

Answer: (a)

1. $x_1 = \frac{1}{\sqrt{2}}\sin\omega t + \frac{1}{\sqrt{2}}\cos\omega t = \sin\left(\omega t + \frac{\pi}{4}\right)$.
2. $x_2 = \sin\omega t + \cos\omega t = \sqrt{2}\sin\left(\omega t + \frac{\pi}{4}\right)$.
3. Both have the same phase $\omega t + \frac{\pi}{4}$; only the amplitudes differ.
4. So the answer is (a) zero.