Question

A particle executes simple harmonic motion of peroid $T$ 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 prependicular to its length. Initially, both the particle and the rod are in their mean positions. The path traced out by the particle will be:

• A. a circle of radius $l$
• B. a straight line inclined at $\frac{\pi }{4}$ to the rod
• C. an ellipse
• D. a figure of eight

Answer 1

The correct option is B a straight line inclined at $\frac{\pi }{4}$ to the rod

Since the time period for SHM of both rod and particle with respect to rod is same, and both start together from mean position, the displacements at any time just get added up.
Since the amplitude of both SHMs is same, the particle goes along the rod from mean position as much as it does perpendicular to it at all times, thus the motion is at an angle of ${45}^{\circ }$ to the rod.