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A particle executes simple harmonic motion of period T and amplitude l along a rod AB of length 2l. The rod AB itself executes simple harmonic motion of the same period and amplitude in a direction perpendicular to its length. Initially, both the particle and the rod are in their mean positions. The path traced out by the particle will be

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a circle of radius l

an ellipse

a straight line inclined at $\large \frac{\pi }{4}$ to the rod

a figure of eight

answer is B.

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Detailed Solution

$\large x\, = \,A\sin \omega t;\,y\,\, = \,\,A\sin \omega t\,;\,x = \,y$

Equation of Straight line , with slope

$\large m\, = \,\tan \theta \, = \,1\, \Rightarrow \,\theta \, = \,{45^o}\, = \,\frac{\pi }{4}$

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