Q. Two linear SHMs of equal amplitude A and angular frequencies $\omega\ and\ 2\omega$ are impressed on a particle along the axes $x\ and\ y$ recpectively. If the initial phase difference between them is $\pi/2,$ the resultant path followed by the particle is
A) $y^{2} = x^{2}(1 - x^{2}/A^{2})$
B) $y^{2} = 2x^{2}(1 - x^{2}A^{2})$
C) $y^{2} = 4x^{2}(1 - x^{2}/A^{2})$
D) $y^{2} = 8x^{2}(1 - x^{2}/A^{2})$
Explanation
(c) $x = A\sin{(\omega + \pi/2)} = A\cos{\omega t}$
$\therefore\ \ \cos{\omega t} = x/A$ and $\sin{\omega t} = \sqrt{1 - (x^{2}/A^{2})}$
$y = A\sin{2\omega t = \sqrt{1 - (x^{2}/\mathrm{\Delta}A^{2})}}$
$y = A\sin{2\omega t} = 2A\sin{\omega t}\cos{\omega t}$
or $y^{2} = 4\ A^{2}\sin^{2}{\omega t}\cos^{2}{\omega t}$
$= 4A^{2} \times \frac{x^{2}}{A^{2}} \times \left( \frac{A^{2} - x^{2}}{A^{2}} \right) = 4x^{2}\left( \begin{array}{r}
1 - \frac{x^{2}}{A^{2}} \\
\
\end{array} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.