Two simple harmonic motions are given by x_{1 | vedclass

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(A) For the first $SHM$: $x_{1} = a \sin \omega t + a \cos \omega t = \sqrt{2} a \sin(\omega t + \frac{\pi}{4})$.Amplitude $A_{1} = \sqrt{2} a$ and ph

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$\sqrt{\frac{3}{2}}$ and $\frac{\pi}{12}$

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(A) For the first $SHM$: $x_{1} = a \sin \omega t + a \cos \omega t = \sqrt{2} a \sin(\omega t + \frac{\pi}{4})$.Amplitude $A_{1} = \sqrt{2} a$ and phase $\phi_{1} = \frac{\pi}{4}$.For the second $SHM$: $x_{2} = a \sin \omega t + \frac{a}{\sqrt{3}} \cos \omega t$.Amplitude $A_{2} = \sqrt{a^{2} + (\frac{a}{\sqrt{3}})^{2}} = \sqrt{a^{2} + \frac{a^{2}}{3}} = \sqrt{\frac{4a^{2}}{3}} = \frac{2a}{\sqrt{3}}$.Phase $\phi_{2} = 	an^{-1}(\frac{a/\sqrt{3}}{a}) = 	an^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$.Ratio of amplitudes $\frac{A_{1}}{A_{2}} = \frac{\sqrt{2} a}{2a/\sqrt{3}} = \frac{\sqrt{2} \cdot \sqrt{3}}{2} = \frac{\sqrt{6}}{2} = \sqrt{\frac{6}{4}} = \sqrt{\frac{3}{2}}$.Phase difference $\Delta \phi = \phi_{1} - \phi_{2} = \frac{\pi}{4} - \frac{\pi}{6} = \frac{3\pi - 2\pi}{12} = \frac{\pi}{12}$.

Two simple harmonic motions are given by $x_{1} = a \sin \omega t + a \cos \omega t$ and $x_{2} = a \sin \omega t + \frac{a}{\sqrt{3}} \cos \omega t$. The ratio of the amplitudes of the first and second motion and the phase difference between them are respectively:

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