Two waves represented by y_1 = a sin frac{2pi | vedclass

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(D) The first wave is $y_1 = a \sin \frac{2\pi}{\lambda} (vt - x)$.The second wave is $y_2 = a \cos \frac{2\pi}{\lambda} (vt - x) = a \sin \left( \fra

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$a\sqrt{2}$

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(D) The first wave is $y_1 = a \sin \frac{2\pi}{\lambda} (vt - x)$.The second wave is $y_2 = a \cos \frac{2\pi}{\lambda} (vt - x) = a \sin \left( \frac{2\pi}{\lambda} (vt - x) + \frac{\pi}{2} \right)$.Comparing these with the standard form $y = A \sin(\omega t - kx + \phi)$,the phase difference between the two waves is $\Delta \phi = \frac{\pi}{2}$.The resultant amplitude $A_{res}$ of two waves with equal amplitude $a$ and phase difference $\Delta \phi$ is given by $A_{res} = \sqrt{a^2 + a^2 + 2a^2 \cos(\Delta \phi)}$.Substituting $\Delta \phi = 90^{\circ}$,we get $A_{res} = \sqrt{a^2 + a^2 + 2a^2 \cos(90^{\circ})} = \sqrt{2a^2 + 0} = a\sqrt{2}$.

Two waves represented by $y_1 = a \sin \frac{2\pi}{\lambda} (vt - x)$ and $y_2 = a \cos \frac{2\pi}{\lambda} (vt - x)$ are superposed. The resultant wave has an amplitude equal to

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