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

A particle executes SHM of period T and amplitude l along a rod AB of

1.	A particle executes SHM of period T and amplitude l along a rod AB of length 2l. The rod AB itself executes SHM of the same period and amplitude in a direction perpendicular to its length. Initially, both the particle and the rod are in their mean position. The path traced out by the particle will be...........
Answer» a circle of radius L a straight line inclined at `(pi)/(4)` to the rod an ellipse a figure of eight Solution :LET the SIMPLE harmonic equation for the particle be `x= l sin omega t"""………."(1)` The equation of motion of rod AB, `y= l cos (omega t + PHI)` where rod executes angular frequency (periodic time) in y-direction with amplitude of particle. For both, at mean position `t=0" and "x= y=0` then `phi = (pi)/(2)`. `therefore y= l cos (omega t+(pi)/(2))= -l sin omega t"..........."(2)` Frqom equation (1) and (2), `y= -x`, which is the equation to a straight line at angle `(pi)/(4)` to the rod.