The displacement equation of a simple harmoni | vedclass

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(C) The given displacement equation is $y = A \sin \omega t - B \cos \omega t$.To find the amplitude,we express the equation in the form $y = R \sin(\

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

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(C) The given displacement equation is $y = A \sin \omega t - B \cos \omega t$.To find the amplitude,we express the equation in the form $y = R \sin(\omega t - \phi)$.Let $A = R \cos \phi$ and $B = R \sin \phi$.Then,$y = R \cos \phi \sin \omega t - R \sin \phi \cos \omega t = R \sin(\omega t - \phi)$.Squaring and adding the expressions for $A$ and $B$:$A^2 + B^2 = R^2 \cos^2 \phi + R^2 \sin^2 \phi = R^2(\cos^2 \phi + \sin^2 \phi) = R^2$.Therefore,the amplitude $R = \sqrt{A^2 + B^2}$.

The displacement equation of a simple harmonic oscillator is given by $y = A \sin \omega t - B \cos \omega t$. The amplitude of the oscillator will be

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