The equations of two waves acting in perpendicular directions are given as\ $x=a \cos (\omega t+\de

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The equations of two waves acting in perpendicular directions are given as\ $x=a \cos (\omega t+\delta)$ and $y=a \cos (\omega t+\alpha)$, where $\delta=\alpha+\frac{\pi}{2}$, the resultant wave represents\

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(b) : Given : $x=a \cos (\omega t+\delta)$\ and $y=a \cos (\omega t+\alpha)$\ where, $\delta=\alpha+\pi / 2$\ $	herefore \quad x=a \cos (\omega t+\alpha+\pi / 2)=-a \sin (\omega t+\alpha)$\ Given the two waves are acting in perpendicular direction with\ the same frequency and phase difference $\pi / 2$.\ From equations (i) and (ii), $$ x^{2}+y^{2}=a^{2} $$ which represents the equation of a circle.\