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(C) Given the displacement equation: $x = a \sin \omega t + b \cos \omega t$. To simplify this,we multiply and divide by $\sqrt{a^2 + b^2}$: $x = \sqr

Vedclass · Accepted Answer

Motion is simple harmonic as well as periodic

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(C) Given the displacement equation: $x = a \sin \omega t + b \cos \omega t$. To simplify this,we multiply and divide by $\sqrt{a^2 + b^2}$: $x = \sqrt{a^2 + b^2} \left( \frac{a}{\sqrt{a^2 + b^2}} \sin \omega t + \frac{b}{\sqrt{a^2 + b^2}} \cos \omega t \right)$. Let $\cos \phi = \frac{a}{\sqrt{a^2 + b^2}}$ and $\sin \phi = \frac{b}{\sqrt{a^2 + b^2}}$. Then the equation becomes: $x = \sqrt{a^2 + b^2} (\sin \omega t \cos \phi + \cos \omega t \sin \phi)$. Using the trigonometric identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we get: $x = A' \sin(\omega t + \phi)$,where $A' = \sqrt{a^2 + b^2}$ is the resultant amplitude. Since the displacement is expressed in the form $x = A' \sin(\omega t + \phi)$,the motion is simple harmonic. Every simple harmonic motion is also periodic. Therefore,the motion is simple harmonic as well as periodic.

The displacement of an oscillator is given by $x = a \sin \omega t + b \cos \omega t$,where $a, b$ and $\omega$ are constants. Then:

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