Q 11-13-018NEETJEE MainMedium
The displacement of a particle is $x = 3\sin\omega t + 4\cos\omega t$ cm. Its motion is
Answer: (D) SHM of amplitude $5$ cm
Two SHMs of the same frequency, $90°$ out of phase, combine into a single SHM of the same frequency with amplitude
$$A = \sqrt{3^2 + 4^2} = 5\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics