The motion of the particle is given by the eq | 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 standard form of simple harmonic motion $x = R \sin(

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simple harmonic with amplitude $(A^2+B^2)^{1/2}$

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(C) Given the equation of motion: $x = A \sin \omega t + B \cos \omega t$. To express this in the standard form of simple harmonic motion $x = R \sin(\omega t + \phi)$,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,$x = \sqrt{A^2 + B^2} (\sin \omega t \cos \phi + \cos \omega t \sin \phi)$. Using the trigonometric identity $\sin(a+b) = \sin a \cos b + \cos a \sin b$,we get: $x = \sqrt{A^2 + B^2} \sin(\omega t + \phi)$. This is the standard equation of simple harmonic motion with amplitude $R = \sqrt{A^2 + B^2} = (A^2 + B^2)^{1/2}$.

The motion of the particle is given by the equation $x = A \sin \omega t + B \cos \omega t$. The motion of the particle is:

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