The displacement of a particle performing SHM at any time t is x=A ( t)+B ( t) , the…
Physics · NEET · NTA Exams — Oscillations and Waves
The displacement of a particle performing SHM at any time t is \(x=A \sin (\omega t)+B \cos (\omega t)\), the amplitude of oscillation is
- A
- B
- \(\sqrt{A^{2}+B^{2}}\)
- A + B
Answer
(C) A^ 2 +B^ 2
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