Question

The velocity of a particle moving in the $x-y$ plane is given by $\frac{dx}{dt}=4\pi \sin \left(3\pi t\right)$ and $\frac{dy}{dt}=3\pi \cos \left(3\pi t\right)$. If the particle starts from origin at $t=0$ then the path of the particle is a/an

• A. straight line
• B. hyperbola
• C. ellipse
• D. parabola

Answer 1

The correct option is C ellipse

For motion along $x-$ axis,
$\frac{dx}{dt}=4\pi \sin \left(3\pi t\right)$
$\Rightarrow dx=4\pi \sin \left(3\pi t\right)dt$
On integrating it with limit, $x=0$ to $x$ and $t=0$ to $t$, we get,
$x=\frac{4}{3}[1-\cos (3\pi t\left)\right]$
$\Rightarrow \cos \left(3\pi t\right)=1-\frac{3x}{4}$ .............(1)
For motion along $y-$ axis,
$\frac{dy}{dt}=3\pi \cos \left(3\pi t\right)$
$\Rightarrow dy=3\pi \cos \left(3\pi t\right)dt$
On integrating it with limit, $y=0$ to $y$ and $t=0$ to $t$, we get,
$y=\sin \left(3\pi t\right)$
$\Rightarrow \sin \left(3\pi t\right)=y$ .............(2)
We know that, ${\sin }^2\theta +{\cos }^2\theta =1$.
From (1) and (2),
$y^2+{(1-\frac{3x}{4})}^2=1$
This equation is similar to the equation of an ellipse, $\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$