Question

Two linear SHMs of equal amplitude $A$ and angular frequencies $\omega$ and $2 \omega$ are impressed on a particle along the axes $x$ and $y$ respectively. If the initial phase difference between them is $\pi / 2$, the resultant path followed by the particle is

• A. $y^{2}=x^{2}\left(1-x^{2} / A^{2}\right)$
• B. $y^{2}=2 x^{2}\left(1-x^{2} / A^{2}\right)$
• C. $y^{2}=4 x^{2}\left(1-x^{2} / A^{2}\right)$
• D. $y^{2}=8 x^{2}\left(1-x^{2} / A^{2}\right)$

Answer 1

Answer: (C)

As, $x=A \sin (\omega+\pi / 2)=A \cos \omega t$
$\therefore \cos \omega t=x / A $
and $\sin \omega t=\sqrt{1-\left(x^{2} / A^{2}\right)}$
$y=\,A\, \sin \, 2\, \omega t=\sqrt{1-\left(x^{2} / \Delta A^{2}\right)}$
$y =\,A \,\sin \,2\, \omega t=2 \,A\, \sin \omega t \cos \omega t $
or $ y^{2} =4 A^{2} \sin ^{2} \omega t \cos ^{2} \omega t $
$=4 A^{2} \times \frac{x^{2}}{A^{2}} \times\left(\frac{A^{2}-x^{2}}{A^{2}}\right)$
$=4 x^{2}\left(1-\frac{x^{2}}{A^{2}}\right)$