Question

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

• A. a circle of radius $l$
• B. a straight line inclined at $\frac{\pi}{4}$ to the rod
• C. an ellipse
• D. a figure of eight

Answer 1

Answer: (B)

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 $y=l \cos (\omega t+\phi)$ where $\phi$ 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 $\phi=\pi / 2$ since $x=y=0$ at $t=0$
So, $y=I \cos (\omega t+\pi / 2)=-i \sin \omega t \,\,\,\,\,\ldots$ (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.