The minimum phase difference between two simple harmonic oscillations, $y_{1} = \frac{1}{2}\sin{\omega t + \frac{\sqrt{3}}{2}\cos{\omega t}}$ $y_{2} = \sin{\omega t + \cos{\omega t}}$, is

Question

Dr. Abubacker · Accepted Answer

(b) Here, $y_{1} = \frac{1}{2}\sin{\omega t + \frac{\sqrt{3}}{2}\cos{\omega t}}$
= $\cos{\frac{\pi}{3}\sin{\omega t + \sin{\frac{\pi}{2}\cos{\omega t}}}}$
$	herefore$ $y_{1} = \sin\left( \omega t + \frac{\pi}{3} ight)$
Similarly, $y_{2} = \sqrt{2}\sin\left( \omega t + \frac{\pi}{4} ight)$
$	herefore$ Phase difference $\mathrm{\Delta}\varnothing = \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12}$

Q. The minimum phase difference between two simple harmonic oscillations, $y_{1} = \frac{1}{2}\sin{\omega t + \frac{\sqrt{3}}{2}\cos{\omega t}}$ $y_{2} = \sin{\omega t + \cos{\omega t}}$, is

Explanation

Reviewed by Dr. Abubacker

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