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JEE Mains · Physics · STD 11 - 13. oscillations
For a periodic motion represented by the equation \(Y=\sin \omega t+\cos \omega t\) The amplitude of the motion is
- A \(0.5\)
- B \(\sqrt{2}\)
- C \(9\)
- D \(6\)
Answer & Solution
Correct Answer
(B) \(\sqrt{2}\)
Step-by-step Solution
Detailed explanation
\(y=\sin \omega t+\cos \omega t\) \(y=\sin \omega t+\sin \left(\omega t+\frac{\pi}{2}\right)\) \(\Delta \phi=\frac{\pi}{2}\) \(A _{\text {net }}=\sqrt{1^2+1^2+2 \times 1 \times 1 \times \cos (\Delta \phi)}\) \(A _{\text {met }}=\sqrt{2}\)
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