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...........

Options

A. a circle of radius l
B. a straight line inclined at `(pi)/(4)` to the rod
C. an ellipse
D. a figure of eight

Answer

B

Explanation

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.