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(C) The first wave is $y_1 = a \sin(\omega t + \frac{\pi}{6})$.The second wave is $y_2 = a \cos(\omega t) = a \sin(\omega t + \frac{\pi}{2})$.The phas

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$\sqrt{3} a$

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(C) The first wave is $y_1 = a \sin(\omega t + \frac{\pi}{6})$.The second wave is $y_2 = a \cos(\omega t) = a \sin(\omega t + \frac{\pi}{2})$.The phase difference between the two waves is $\phi = (\omega t + \frac{\pi}{2}) - (\omega t + \frac{\pi}{6}) = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$.The resultant amplitude $A$ of two waves with equal amplitude $a$ is given by $A = \sqrt{a_1^2 + a_2^2 + 2a_1 a_2 \cos \phi}$.Substituting $a_1 = a$,$a_2 = a$,and $\phi = \frac{\pi}{3}$:$A = \sqrt{a^2 + a^2 + 2(a)(a) \cos(\frac{\pi}{3})}$$A = \sqrt{2a^2 + 2a^2(\frac{1}{2})}$$A = \sqrt{2a^2 + a^2} = \sqrt{3a^2} = \sqrt{3} a$.

Two waves are represented by $y_1 = a \sin(\omega t + \frac{\pi}{6})$ and $y_2 = a \cos(\omega t)$. What will be their resultant amplitude?

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