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(B) The resultant displacement is given by $x = x_1 + x_2 + x_3$. Substituting the given expressions: $x = A \cos \omega t + 2 A \sin \omega t + \sqrt

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$\sqrt{5} A$

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(B) The resultant displacement is given by $x = x_1 + x_2 + x_3$. Substituting the given expressions: $x = A \cos \omega t + 2 A \sin \omega t + \sqrt{2} A \cos (\omega t + \frac{\pi}{4})$. Using the trigonometric identity $\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$: $x = A \cos \omega t + 2 A \sin \omega t + \sqrt{2} A (\cos \omega t \cos \frac{\pi}{4} - \sin \omega t \sin \frac{\pi}{4})$. Since $\cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$: $x = A \cos \omega t + 2 A \sin \omega t + \sqrt{2} A (\frac{1}{\sqrt{2}} \cos \omega t - \frac{1}{\sqrt{2}} \sin \omega t)$. $x = A \cos \omega t + 2 A \sin \omega t + A \cos \omega t - A \sin \omega t$. $x = 2 A \cos \omega t + A \sin \omega t$. The resultant amplitude $R$ for a displacement of the form $x = a \cos \omega t + b \sin \omega t$ is $R = \sqrt{a^2 + b^2}$. Therefore,$R = \sqrt{(2 A)^2 + A^2} = \sqrt{4 A^2 + A^2} = \sqrt{5} A$.

The amplitude of the wave resulting from the superposition of $3$ waves given by $x_1 = A \cos \omega t$,$x_2 = 2 A \sin \omega t$ and $x_3 = \sqrt{2} A \cos (\omega t + \frac{\pi}{4})$ is

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