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 sotion 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 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. parabolic

Answer

B

Explanation

Let the simple harmonic equation for the particle be `x=//sin omegat`……..(i)
Where `omega` 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 equilibrium is `y=//cos(omegat+phi)` where `phi` is the intial phase difference(phase angle for y). But both the particle as well as the rod pass thorugh the mean position simultaneously. Hence `phi=pi//2`since x=y=o at t=0.
So, `y=//cos(omegat+pi//2)=-//sinomegat`......(ii)
Eliminating | from equation (i)and (ii), we have `y=-x` or y=x, which is the equation to a straight line at angle `pi//4` to the rod.