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

Answer 1

Answer: (C)

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