Two SHM are represented by equations,y_1 = 6c | vedclass

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(A) Given equations are:$y_1 = 6 \cos \left( 6 \pi t + \frac{\pi}{6} \right)$$y_2 = 3 \left( \sqrt{3} \sin 3 \pi t + \cos 3 \pi t \right)$For $y_1$,th

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The ratio of their amplitudes is $1$.

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(A) Given equations are:$y_1 = 6 \cos \left( 6 \pi t + \frac{\pi}{6} \right)$$y_2 = 3 \left( \sqrt{3} \sin 3 \pi t + \cos 3 \pi t \right)$For $y_1$,the amplitude $A_1 = 6$.For $y_2$,we can rewrite it as:$y_2 = 6 \left( \frac{\sqrt{3}}{2} \sin 3 \pi t + \frac{1}{2} \cos 3 \pi t \right)$Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we set $\cos \phi = \frac{\sqrt{3}}{2}$ and $\sin \phi = \frac{1}{2}$,which gives $\phi = \frac{\pi}{6}$.$y_2 = 6 \sin \left( 3 \pi t + \frac{\pi}{6} \right)$Thus,the amplitude $A_2 = 6$.The ratio of their amplitudes is $\frac{A_1}{A_2} = \frac{6}{6} = 1$.Therefore,option $A$ is the correct answer.

Two $SHM$ are represented by equations,$y_1 = 6\cos \left( {6\pi t + \frac{\pi }{6}} \right)$ and $y_2 = 3\left( {\sqrt 3 \sin 3\pi t + \cos 3\pi t} \right)$. Which of the following statements is true?

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