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(D) Given equations are $y_1 = a \sin \omega t$ and $y_2 = a \cos \omega t$.We can rewrite $y_2$ using the trigonometric identity $\cos 	heta = \sin(

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Lags the second by $\frac{\pi}{2}$

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(D) Given equations are $y_1 = a \sin \omega t$ and $y_2 = a \cos \omega t$.We can rewrite $y_2$ using the trigonometric identity $\cos 	heta = \sin(	heta + \frac{\pi}{2})$.So,$y_2 = a \sin(\omega t + \frac{\pi}{2})$.Comparing the phases,the phase of $y_1$ is $\omega t$ and the phase of $y_2$ is $\omega t + \frac{\pi}{2}$.The phase difference is $\phi = (\omega t + \frac{\pi}{2}) - \omega t = \frac{\pi}{2}$.Since $y_2$ has a phase that is $\frac{\pi}{2}$ greater than $y_1$,$y_1$ lags behind $y_2$ by $\frac{\pi}{2}$.

Two waves are represented by the equations $y_1 = a \sin \omega t$ and $y_2 = a \cos \omega t$. The first wave

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