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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
- A \(\mathrm{A}_{0}+\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}}\)
- B \(\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}}\)
- C \(\sqrt{\mathrm{A}_{0}^{2}+(\mathrm{A}+\mathrm{B})^{2}}\)
- D \(\mathrm{A}+\mathrm{B}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}}\)
Step-by-step Solution
Detailed explanation
\(y=A_{0}+A \sin \omega t+B \cos \omega t\) \(y=A_{0}+\sqrt{A^{2}+B^{2}} \sin (\omega t+\phi)\)
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