Two simple harmonic motions are represented b | vedclass

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(B) The first equation is $x_{1} = 5 \sin(2 \pi t + \frac{\pi}{4})$. The amplitude $A_{1} = 5$.The second equation is $x_{2} = 5 \sqrt{2}(\sin 2 \pi t

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$2$

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(B) The first equation is $x_{1} = 5 \sin(2 \pi t + \frac{\pi}{4})$. The amplitude $A_{1} = 5$.The second equation is $x_{2} = 5 \sqrt{2}(\sin 2 \pi t + \cos 2 \pi t)$.To find the amplitude,we rewrite the expression inside the parentheses by multiplying and dividing by $\sqrt{2}$:$x_{2} = 5 \sqrt{2} \cdot \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin 2 \pi t + \frac{1}{\sqrt{2}} \cos 2 \pi t \right)$$x_{2} = 10 \left( \sin 2 \pi t \cos \frac{\pi}{4} + \cos 2 \pi t \sin \frac{\pi}{4} \right)$Using the identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we get:$x_{2} = 10 \sin(2 \pi t + \frac{\pi}{4})$.The amplitude $A_{2} = 10$.Therefore,the ratio of the amplitudes is $\frac{A_{2}}{A_{1}} = \frac{10}{5} = 2$.

Two simple harmonic motions are represented by the equations $x_{1} = 5 \sin(2 \pi t + \frac{\pi}{4})$ and $x_{2} = 5 \sqrt{2}(\sin 2 \pi t + \cos 2 \pi t)$. The amplitude of the second motion is ....... times the amplitude of the first motion.

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