If x = a sin left( omega t + frac{pi}{6} righ | vedclass

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(A) Given the two wave equations:$x = a \sin \left( \omega t + \frac{\pi}{6} \right)$$x' = a \cos \omega t$To find the phase difference,we express bot

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$ \pi/3 $

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(A) Given the two wave equations:$x = a \sin \left( \omega t + \frac{\pi}{6} ight)$$x' = a \cos \omega t$To find the phase difference,we express both equations in terms of the sine function.Using the identity $\cos 	heta = \sin \left( 	heta + \frac{\pi}{2} ight)$,we can rewrite $x'$ as:$x' = a \sin \left( \omega t + \frac{\pi}{2} ight)$Now,the phase of the first wave is $\phi_1 = \omega t + \frac{\pi}{6}$ and the phase of the second wave is $\phi_2 = \omega t + \frac{\pi}{2}$.The phase difference $\Delta \phi$ is given by:$\Delta \phi = \phi_2 - \phi_1 = \left( \omega t + \frac{\pi}{2} ight) - \left( \omega t + \frac{\pi}{6} ight)$$\Delta \phi = \frac{\pi}{2} - \frac{\pi}{6} = \frac{3\pi - \pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}$Thus,the phase difference between the two waves is $\frac{\pi}{3}$.

If $x = a \sin \left( \omega t + \frac{\pi}{6} \right)$ and $x' = a \cos \omega t$,then what is the phase difference between the two waves?

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