Question

A particle of mass $m$ moves according to the equation $F=-amr$ where $a$ is a positive constant$,r$ is radius vector. $r=r_0\hat{i}$ and $v=v_0\hat{j}$ at $t=0$. Describe the trajectory.

• A. ${\left(\frac{x}{r_0}\right)}^2+a{\left(\frac{y}{v_0}\right)}^2=1$
• B. ${\left(\frac{x}{r_0}\right)}^2+a{\left(\frac{y}{v_0}\right)}^2=0$
• C. ${\left(\frac{x}{r_0}\right)}^2+{\left(\frac{y}{v_0}\right)}^2=\frac{1}{\alpha }$
• D. $noneofthese$

Answer 1

The correct option is A ${\left(\frac{x}{r_0}\right)}^2+a{\left(\frac{y}{v_0}\right)}^2=1$

$F=-amr\Rightarrow F_x\hat{i}+F_y\hat{j}=-am(x\hat{i}+y\hat{j})\Rightarrow m\frac{d^{2}x}{d{t}^2}\hat{i}+m\frac{d^{2}y}{d{t}^2}\hat{j}=-am(x\hat{i}+y\hat{j})$
$\frac{d^{2}x}{d{t}^2}=-ax$ and $\frac{d^{2}y}{d{t}^2}=-ay$ ...............................................................$\left(i\right)$
Thus $x=x_o\sin (\omega t+\theta )$ and ${\omega }^2=a$ .........................................................................................$\left(ii\right)$
$\frac{dx}{dt}=x_o\omega \cos (\omega t+\theta )$
since $v_x=0$ when $t=0$
$\Rightarrow x_o\omega \cos \left(\theta \right)=0$
$\Rightarrow \theta =\frac{\pi }{2}$
$\Rightarrow x=x_o\sin (\omega t+\frac{\pi }{2})$
$x=x_o\cos \left(\omega t\right)$
$\Rightarrow x=r_o\cos \left(\omega t\right)$ ...............................................$\left(iii\right)$ $\because x(t=0)=r_o$
Now from $\left(i\right)$, we have,
$y=y_0\sin (\omega t+\beta )$
$y=y_0\sin \left(\omega t\right)\because y(t=0)=0$
$\frac{dy}{dt}=y_0\omega \cos \left(\omega t\right)$
since $v_y=v_o$ when $t=0$
$\Rightarrow v_o=y_0\omega$
$\Rightarrow y=\frac{v_o}{\omega }\sin \left(\omega t\right)$ .....$\left(iv\right)$
from $\left(iii\right)$ and $\left(iv\right)$ we have,
${\left(\frac{x}{r_0}\right)}^2+{\left(\frac{y}{v_0/\omega }\right)}^2=1$
$\Rightarrow {\left(\frac{x}{r_0}\right)}^2+{\omega }^2{\left(\frac{y}{v_0}\right)}^2=1$
$\Rightarrow {\left(\frac{x}{r_0}\right)}^2+a{\left(\frac{y}{v_0}\right)}^2=1\because {\omega }^2=a$ from $\left(ii\right)$