Two waves are represented as y1=2 a sin(ωt+π/ 6) and y2=-2 a cos(ωt-(π)/(6)). The phase differen

step 1 In this question we have given the equation of two waves and we have to find the face difference

Two waves are represented as y1=2 a sin(ωt+π/ 6) and y2=-2 a...

Two waves are represented as $y_{1}=2 a \sin (\omega t+\pi / 6)$ and $y_{2}=-2 a \cos \left(\omega t-\frac{\pi}{6}\right)$. The phase difference between the two waves is
(A) $\frac{\pi}{3}$
(B) $\frac{4 \pi}{3}$
(C) $\frac{3 \pi}{3}$
(D) $\frac{5 \pi}{6}$

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Key Concepts

Phase Difference

This concept involves determining the relative phase offset between two sinusoidal waves. The phase difference is obtained by comparing the phase constants (or the arguments of the sine or cosine functions) of the waves, once they are expressed in a common form. It is a key parameter that impacts the interference and resultant amplitude when two waves meet.

Trigonometric Conversions

This concept refers to using trigonometric identities to convert between sine and cosine functions. For example, the identity cos(x) = sin(x + ?/2) can be applied to represent both waveforms in the same trigonometric form, which simplifies the process of identifying and comparing their respective phases.

Phase Shift Due to Negative Coefficients

This concept involves understanding how multiplying a sinusoidal function by -1 affects its phase. A negative sign can be interpreted as an additional phase shift of ? (or 180°) in the waveform. Recognizing this effect is crucial when aligning the phases of waves to accurately calculate the phase difference.

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