ExamBro
ExamBro
NEET · Physics · STD 11 - 13. oscillations

The displacement of a particle executing simple harmonic motion is given by \(\mathrm{y}=\mathrm{A}_{0}+\mathrm{A} \sin \omega \mathrm{t}+\mathrm{B} \cos \omega \mathrm{t}\) Then the amplitude of its oscillation is given by

  1. A \(\mathrm{A}_{0}+\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}}\)
  2. B \(\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}}\)
  3. C \(\sqrt{\mathrm{A}_{0}^{2}+(\mathrm{A}+\mathrm{B})^{2}}\)
  4. D \(\mathrm{A}+\mathrm{B}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}}\)

Step-by-step Solution

Detailed explanation

\(y=A_{0}+A \sin \omega t+B \cos \omega t\) \(y=A_{0}+\sqrt{A^{2}+B^{2}} \sin (\omega t+\phi)\)
Same subject
Explore more questions on app