Oscillations Ques 14
14. 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: 14.(d)
Solution:
- (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.
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