The motion of a particle as per x = A sin ome | vedclass

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(C) Given the equation of motion: $x = A \sin \omega t + B \cos \omega t$.To express this in the form of $x = R \sin(\omega t + \phi)$,we multiply and

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$SHM$ with amplitude $\sqrt{A^2 + B^2}$

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(C) Given the equation of motion: $x = A \sin \omega t + B \cos \omega t$.To express this in the form of $x = R \sin(\omega t + \phi)$,we multiply and divide by $\sqrt{A^2 + B^2}$:$x = \sqrt{A^2 + B^2} \left[ \frac{A}{\sqrt{A^2 + B^2}} \sin \omega t + \frac{B}{\sqrt{A^2 + B^2}} \cos \omega t \right]$.Let $\frac{A}{\sqrt{A^2 + B^2}} = \cos \phi$ and $\frac{B}{\sqrt{A^2 + B^2}} = \sin \phi$.Then the equation becomes $x = \sqrt{A^2 + B^2} (\sin \omega t \cos \phi + \cos \omega t \sin \phi)$.Using the trigonometric identity $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$,we get:$x = \sqrt{A^2 + B^2} \sin(\omega t + \phi)$.This represents a Simple Harmonic Motion $(SHM)$ with amplitude $R = \sqrt{A^2 + B^2}$.

The motion of a particle as per $x = A \sin \omega t + B \cos \omega t$ is :-

Similar Questions

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