What is the phase difference between two simp | vedclass

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(B) Given equations for simple harmonic motion are:$x_{1}=A \sin \left(\omega t+\frac{\pi}{6}\right)$$x_{2}=A \cos (\omega t)$To find the phase differ

Vedclass · Accepted Answer

$\frac{\pi}{3}$

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(B) Given equations for simple harmonic motion are:$x_{1}=A \sin \left(\omega t+\frac{\pi}{6}ight)$$x_{2}=A \cos (\omega t)$To find the phase difference,we must express both equations in the same trigonometric function (sine).Using the identity $\cos(	heta) = \sin(	heta + \frac{\pi}{2})$,we can rewrite $x_{2}$ as:$x_{2}=A \sin \left(\omega t+\frac{\pi}{2}ight)$Now,the phase of the first motion is $\phi_{1} = \omega t + \frac{\pi}{6}$ and the phase of the second motion is $\phi_{2} = \omega t + \frac{\pi}{2}$.The phase difference $\Delta \phi$ is given by:$\Delta \phi = \phi_{2} - \phi_{1}$$\Delta \phi = \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}$Therefore,the phase difference is $\frac{\pi}{3}$.

What is the phase difference between two simple harmonic motions represented by $x_{1}=A \sin \left(\omega t+\frac{\pi}{6}\right)$ and $x_{2}=A \cos (\omega t)$?

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