Two simple harmonic motions are represented b | vedclass

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(B) The first equation is $y_{1} = 5[\sin 2 \pi t + \sqrt{3} \cos 2 \pi t]$.We can rewrite this by multiplying and dividing by $2$: $y_{1} = 10 [\frac

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$2: 1$

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(B) The first equation is $y_{1} = 5[\sin 2 \pi t + \sqrt{3} \cos 2 \pi t]$.We can rewrite this by multiplying and dividing by $2$: $y_{1} = 10 [\frac{1}{2} \sin 2 \pi t + \frac{\sqrt{3}}{2} \cos 2 \pi t]$.Using the trigonometric identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we set $\cos \frac{\pi}{3} = \frac{1}{2}$ and $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$.Thus,$y_{1} = 10 [\sin 2 \pi t \cos \frac{\pi}{3} + \cos 2 \pi t \sin \frac{\pi}{3}] = 10 \sin(2 \pi t + \frac{\pi}{3})$.Therefore,the amplitude $A_{1} = 10$.For the second equation,$y_{2} = 5 \sin (2 \pi t + \frac{\pi}{4})$,the amplitude $A_{2} = 5$.The ratio of the amplitudes is $\frac{A_{1}}{A_{2}} = \frac{10}{5} = \frac{2}{1}$.

Two simple harmonic motions are represented by $y_{1} = 5[\sin 2 \pi t + \sqrt{3} \cos 2 \pi t]$ and $y_{2} = 5 \sin (2 \pi t + \frac{\pi}{4})$. The ratio of their amplitudes is

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