Oscillations - Result Question 16

19. Two simple harmonic motions act on a particle. These harmonic motions are $x=A \cos (\omega t+\delta)$, $y=A \cos (\omega t+\alpha)$ when $\delta=\alpha+\frac{\pi}{2}$, the resulting motion is

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(a) a circle and the actual motion is clockwise

(b) an ellipse and the actual motion is counterclockwise

(c) an ellipse and the actual motion is clockwise

(d) a circle and the actual motion is counter clockwise

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Answer:

Correct Answer: 19. (d)

Solution:

  1. (d) $x=A \cos (\omega t+\delta)$

$y=A \cos (\omega t+\alpha)$

When $\delta=\alpha+\frac{\pi}{2}$

$x=A \cos (\frac{\pi}{2}+\omega t+\alpha)$

$x=-A \sin (\omega t+\alpha)$

Squaring (1) and (2) and then adding $x^{2}+y^{2}=A^{2}[\cos ^{2}(\omega t+\alpha)+\sin ^{2}(\omega t+\alpha)]$ or $x^{2}+y^{2}=A^{2}$, which is the equation of a circle. The present motion is anticlockwise.