The minimum phase difference between two simp | vedclass

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(A) Given equations are:$x_1 = \frac{1}{\sqrt{2}} \sin \omega t + \frac{1}{\sqrt{2}} \cos \omega t$Using the identity $\sin(A+B) = \sin A \cos B + \co

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(A) Given equations are:$x_1 = \frac{1}{\sqrt{2}} \sin \omega t + \frac{1}{\sqrt{2}} \cos \omega t$Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$, we can write:$x_1 = \sin \omega t \cos \frac{\pi}{4} + \cos \omega t \sin \frac{\pi}{4} = \sin(\omega t + \frac{\pi}{4})$Now, for $x_2 = \sin \omega t + \cos \omega t$, multiply and divide by $\sqrt{2}$:$x_2 = \sqrt{2} (\frac{1}{\sqrt{2}} \sin \omega t + \frac{1}{\sqrt{2}} \cos \omega t) = \sqrt{2} \sin(\omega t + \frac{\pi}{4})$Comparing the phases of $x_1$ and $x_2$, both have the same phase $\phi = \omega t + \frac{\pi}{4}$.Therefore, the phase difference $\Delta \phi = (\omega t + \frac{\pi}{4}) - (\omega t + \frac{\pi}{4}) = 0$.

The minimum phase difference between two simple harmonic motions $x_1 = \frac{1}{\sqrt{2}} \sin \omega t + \frac{1}{\sqrt{2}} \cos \omega t$ and $x_2 = \sin \omega t + \cos \omega t$ is $[\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}]$

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