91Ó°ÊÓ

First, we find the first derivative (velocity) function, \(v(t)\): \[\begin{aligned} v(t) = \frac{dy}{dt} = a\frac{d(\sin \omega t)}{dt} + b\frac{d(\cos \omega t)}{dt} \end{aligned}\] Now, using the chain rule, we obtain: \[v(t) = a(\omega \cos \omega t) - b(\omega \sin \omega t)\] Next, we find the second derivative (acceleration) function, \(a(t)\): \[\begin{aligned} a(t) = \frac{dv}{dt} = a\frac{d(\omega \cos \omega t)}{dt} - b\frac{d(\omega \sin \omega t)}{dt} \end{aligned}\] Now, using the chain rule again, we obtain: \[a(t) = -a(\omega^2 \sin \omega t) - b(\omega^2 \cos \omega t)\]

To determine whether the motion is SHM, we need to check if the acceleration function is proportional to the displacement function multiplied by a negative constant. In other words, we need to see if: \[a(t) = -k(y)\] Using our previous results, we can rewrite this equation as: \[-a(\omega^2 \sin \omega t) - b(\omega^2 \cos \omega t)=-k(a \sin \omega t+b \cos \omega t)\] Comparing the terms, we find that: \[-a \omega^2 = -ka\] and \[-b \omega^2 = -kb\] Solving these equations for \(k\), we find that \(k = \omega^2\). Since both equations give us the same \(k\) and every term matches, this indicates that the motion is indeed Simple Harmonic Motion.

To find the amplitude of the SHM, we will use the following formula for the displacement of a particle in an SHM: \[y = A \sin(\omega t + \phi)\] where \(A\) is the amplitude, and \(\phi\) is the phase angle, which can vary depending on the initial conditions. Comparing this equation to our original displacement function, we can rewrite our original function as: \[y = A \sin(\omega t + \phi)\] where \[A\sin\phi = a, \, A\cos\phi = b\] To find the amplitude \(A\), we can square the above two equations and add them. We get \[A^2 \sin^2\phi + A^2 \cos^2\phi = a^2 + b^2\] \[A^2(\sin^2\phi + \cos^2\phi) = a^2 + b^2\] Since \(\sin^2\phi + \cos^2\phi = 1\), we find that the amplitude of the SHM is: \[A = \sqrt{a^2 + b^2}\] Therefore, the displacement function describes a Simple Harmonic Motion with amplitude \(\sqrt{a^2+b^2}\). The correct option is (D).