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Superposition of Perpendicular SHMs
Concepts tested here
- Equal amplitudes with quarter-period phase difference
All Questions
2000 AIPMT 1 question
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The equations of two waves are given as $x = a\cos(\omega t + \delta)$ and $y = a\cos(\omega t + \alpha)$, where $\delta = \alpha + \dfrac{\pi}{2}$, then resultant wave represents :Take $\theta = \omega t + \alpha$. Then $y = a\cos\theta$ and $x = a\cos\left(\theta + \dfrac{\pi}{2}\right) = -a\sin\theta$.
$x^2 + y^2 = a^2$: the resultant path is a circle of radius a (equal amplitudes and phase difference $\dfrac{\pi}{2}$).
Direction: at $\theta = 0$, the point is at (0, a). A little later ($\theta$ slightly positive), $x = -a\sin\theta$ is negative and y is still nearly a, so the point moves from the top towards the left.
Moving from the positive y-axis towards the negative x-axis is anticlockwise motion.
So the resultant represents a circle traced anticlockwise (a.c.w).