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

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(B) The given equation is $x = 2A \cos \omega t + A \cos (\omega t + \frac{\pi}{2}) + A \cos (\omega t + \pi) + \frac{A}{2} \cos (\omega t + \frac{3\p

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

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(B) The given equation is $x = 2A \cos \omega t + A \cos (\omega t + \frac{\pi}{2}) + A \cos (\omega t + \pi) + \frac{A}{2} \cos (\omega t + \frac{3\pi}{2})$.Using trigonometric identities $\cos(\omega t + \frac{\pi}{2}) = -\sin \omega t$,$\cos(\omega t + \pi) = -\cos \omega t$,and $\cos(\omega t + \frac{3\pi}{2}) = \sin \omega t$,we get:$x = 2A \cos \omega t - A \sin \omega t - A \cos \omega t + \frac{A}{2} \sin \omega t$$x = (2A - A) \cos \omega t + (-A + \frac{A}{2}) \sin \omega t$$x = A \cos \omega t - \frac{A}{2} \sin \omega t$This is of the form $x = A_{res} \cos(\omega t + \phi)$,where the resultant amplitude $A_{res} = \sqrt{A^2 + (-\frac{A}{2})^2}$.$A_{res} = \sqrt{A^2 + \frac{A^2}{4}} = \sqrt{\frac{5A^2}{4}} = \frac{\sqrt{5}A}{2}$.

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

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