Question

A particle of mass $m$ oscillates along $x$-axis according to equation $x=a\sin \omega$ . The nature of the graph between momentum and displacement of the particle is

• A. Circle
• B. Hyperbola
• C. Ellipse
• D. Straight line passing through origin

Answer 1

Answer: (C)

$x=a\sin \omega t$ or $\frac{x}{a}=\sin \omega t\dots$ (i)
Velocity, $v=\frac{dx}{dt}=a\omega \cos \omega t\dots$ (ii)
$\frac{v}{a\omega }=\cos \omega t$
Squaring and adding (i) and (ii), we get
$\frac{x^2}{a^2}+\frac{v^2}{a^2{\omega }^2}={\sin }^2\omega t+{\cos }^2\omega t$
$\frac{x^2}{a^2}+\frac{y^2}{a^2{\omega }^2}=1$
It is an equation of ellipse.
Hence, the graph between velocity and displacement is an ellipse.
Momentum of the particle $=mv$
$\therefore$ The nature of graph of the momentum and displacement is same as that of velocity and displacement.