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(B) The given equation is $Y = \sin \omega t + \cos \omega t$.We can rewrite $\cos \omega t$ as $\sin(\omega t + \frac{\pi}{2})$.So,$Y = \sin \omega t

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$\sqrt{2}$

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(B) The given equation is $Y = \sin \omega t + \cos \omega t$.We can rewrite $\cos \omega t$ as $\sin(\omega t + \frac{\pi}{2})$.So,$Y = \sin \omega t + \sin(\omega t + \frac{\pi}{2})$.This represents the superposition of two simple harmonic motions with amplitudes $A_1 = 1$ and $A_2 = 1$,and a phase difference $\Delta \phi = \frac{\pi}{2}$.The resultant amplitude $A_{	ext{net}}$ is given by the formula $A_{	ext{net}} = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\Delta \phi)}$.Substituting the values: $A_{	ext{net}} = \sqrt{1^2 + 1^2 + 2(1)(1) \cos(\frac{\pi}{2})}$.Since $\cos(\frac{\pi}{2}) = 0$,we get $A_{	ext{net}} = \sqrt{1 + 1} = \sqrt{2}$.

For a periodic motion represented by the equation $Y = \sin \omega t + \cos \omega t$,the amplitude of the motion is:

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