The displacement of a particle executing simp | vedclass

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(B) The given equation is $y = A_{0} + A \sin \omega t + B \cos \omega t$.To find the amplitude,we express the time-dependent part in the form $R \sin

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$\sqrt{A^{2} + B^{2}}$

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(B) The given equation is $y = A_{0} + A \sin \omega t + B \cos \omega t$.To find the amplitude,we express the time-dependent part in the form $R \sin(\omega t + \phi)$.Let $A = R \cos \phi$ and $B = R \sin \phi$.Then $A^{2} + B^{2} = R^{2} (\cos^{2} \phi + \sin^{2} \phi) = R^{2}$.Thus,$R = \sqrt{A^{2} + B^{2}}$.The equation becomes $y = A_{0} + \sqrt{A^{2} + B^{2}} \sin(\omega t + \phi)$.Here,$A_{0}$ represents the shift in the mean position,and the coefficient of the trigonometric term,$\sqrt{A^{2} + B^{2}}$,represents the amplitude of the oscillation.

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

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