Physics - SHM Question with Solution | TestHub

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 $y=l\cos (\omega t+\varphi )$ where $\varphi$ 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 $\varphi =\pi /2$ since $x=y=0$ at $t=0$.
Eliminating $t$ between (i) and (ii), we have $y=-x$ which is the equation of a straight line at angle $\pi /4$ to the rod.