To find the amplitude of the oscillation for a particle executing simple harmonic motion, we start with the given displacement equation:$\\ y = A_0 + A \sin \omega t + B \cos \omega t \\$In simple harmonic motion, the displacement can be expressed as:$\\ y = C \sin(\omega t + \phi) \\$where $C$ is the amplitude and $\phi$ is the phase angle.$\\$We need to express $A \sin \omega t + B \cos \omega t$ in the form $C \sin(\omega t + \phi). \\$Using the trigonometric identity:$\\ C \sin(\omega t + \phi) = C (\sin \omega t \cos \phi + \cos \omega t \sin \phi) \\$Comparing coefficients, we have:$\\ A = C \cos \phi \\ B = C \sin \phi \\$To find $C,$ we use the Pythagorean identity:$\\ C^2 = (C \cos \phi)^2 + (C \sin \phi)^2 \\ C^2 = A^2 + B^2 \\ C = \sqrt{A^2 + B^2} \\$Thus, the amplitude of the oscillation is $\sqrt{A^2 + B^2}. \\$Therefore, the correct option is Option 2.