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
Concept. Two SHMs of the same frequency combine into one SHM; $90°$ apart they give $A = \sqrt{A_1^2 + A_2^2}$.
Step 1. Amplitude. $\sqrt{3^2 + 4^2} = 5$ cm.
Answer. SHM of amplitude $5$ cm (Option D).
Solution by Sreeraj P, M.Sc Physics