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 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:

Options

A. a cricle 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" …(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 equation is `y=lcos(omega t+phi)` where `phi` is the inital phase difference (phase angel for y). But both the particle as well as the rod pass through the mean position simultaneously. Hence `phi=pi//2` since x = y = 0 at t = 0.
So, `y=l cos (omega t+pi//2)=-l sin omega t" ...(ii)"`
Eliminating t between (i) and (ii), we have `y=-x`
Which is the equation to a straight line at angle `pi//4` to the rod.