A particle executes simple harmonic motion of period T and amplitude l along a rod A B of length 2l. The rod A B 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

Step-by-step solution

where is its angular velocity. Since the S.H.M. of the rod has the same period and amplitude and its vibration is perpendicular to that of the particle, its equation is y=l ( t+ ) where is the initial phase difference (phase angle for y ). But both the particle as well as the rod pass through the mean position simultaneously. Hence = / 2 since x=y=0 at t=0 . Eliminating t between (i) and (ii), we have y=-x which is the equation of a straight line at angle / 4 to the rod.