Question

x and y displacements of a particle are given as x(t) = a $sin\omega t$ and y(t) = a $sin2\omega t$. Its trajectory will look like :

Answer 1

$x=a\sin \omega t$
$y=a\sin 2\omega t$
$y=\left(2\frac{x}{a}\sqrt{1-\frac{x^2}{a^2}}\right)a$
$y^2=4x^2(1-\frac{x^2}{a^2})$
At $y=u$ when $x=0$
$x=\pm a$
$\sin \omega t=\frac{x}{a}$
$y=a2\sin \omega t\cos \omega t$
$=2\cdot a\sin \omega t\cos \omega t$
$\sin \omega t=\frac{x}{a}$
${\sin }^2\omega t=\frac{x^2}{a^2}$
$1-{\sin }^2\omega t={\cos }^2\omega t$
$1-\frac{x^2}{a^2}={\cos }^2\omega t$
$\cos \omega t=\sqrt{1-\frac{x^2}{a^2}}$
So, $y=a.2\sin \omega t\cos \omega t$
$=2.a\left(\frac{x}{a}\sqrt{1-\frac{x^2}{a^2}}\right)$