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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2023
  • Easy
  • MCQ

For a periodic motion represented by the equation \(Y=\sin \omega t+\cos \omega t\) The amplitude of the motion is 

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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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