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(D) The given equation is $x = 2 \sin \omega t + 4 \cos \omega t$.Any equation of the form $x = A_1 \sin \omega t + A_2 \cos \omega t$ can be rewritte

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$2 \sqrt{5}$

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(D) The given equation is $x = 2 \sin \omega t + 4 \cos \omega t$.Any equation of the form $x = A_1 \sin \omega t + A_2 \cos \omega t$ can be rewritten as $x = A \sin(\omega t + \phi)$,where the amplitude $A$ is given by $A = \sqrt{A_1^2 + A_2^2}$.Here,$A_1 = 2$ and $A_2 = 4$.Substituting these values into the formula:$A = \sqrt{2^2 + 4^2}$$A = \sqrt{4 + 16}$$A = \sqrt{20}$$A = \sqrt{4 	imes 5} = 2 \sqrt{5}$.Therefore,the amplitude of oscillation is $2 \sqrt{5}$ units.

The $S.H.M.$ of a particle is given by the equation $x = 2 \sin \omega t + 4 \cos \omega t$. Its amplitude of oscillation is ........ units.

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