A vibratory motion is represented by x = 2A c | vedclass

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(B) Given the equation: $x = 2A \cos \omega t + A \cos \left( \omega t + \frac{\pi}{2} \right) + A \cos ( \omega t + \pi ) + \frac{A}{2} \cos \left( \

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$\frac{\sqrt{5}A}{2}$

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(B) Given the equation: $x = 2A \cos \omega t + A \cos \left( \omega t + \frac{\pi}{2} \right) + A \cos ( \omega t + \pi ) + \frac{A}{2} \cos \left( \omega t + \frac{3\pi}{2} \right)$.Using trigonometric identities: $\cos ( \omega t + \frac{\pi}{2} ) = -\sin \omega t$,$\cos ( \omega t + \pi ) = -\cos \omega t$,and $\cos ( \omega t + \frac{3\pi}{2} ) = \sin \omega t$.Substituting these into the equation:$x = 2A \cos \omega t - A \sin \omega t - A \cos \omega t + \frac{A}{2} \sin \omega t$.Simplifying the terms:$x = (2A - A) \cos \omega t + (\frac{A}{2} - A) \sin \omega t = A \cos \omega t - \frac{A}{2} \sin \omega t$.The resultant amplitude $A_R$ for a motion of the form $x = a \cos \omega t + b \sin \omega t$ is given by $A_R = \sqrt{a^2 + b^2}$.Here,$a = A$ and $b = -\frac{A}{2}$.$A_R = \sqrt{A^2 + (-\frac{A}{2})^2} = \sqrt{A^2 + \frac{A^2}{4}} = \sqrt{\frac{5A^2}{4}} = \frac{\sqrt{5}A}{2}$.

$A$ vibratory motion is represented by $x = 2A \cos \omega t + A \cos \left( \omega t + \frac{\pi}{2} \right) + A \cos ( \omega t + \pi ) + \frac{A}{2} \cos \left( \omega t + \frac{3\pi}{2} \right)$. The resultant amplitude of the motion is

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