Q114 / 180 · NEET 2013 (Karnataka) · Physics

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

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

Solution

According to the question $$ x=\operatorname{arsin} \omega t \rightarrow \frac{x}{a}=\sin \omega t $$ Now velocity $\mathrm{V}=\frac{d x}{d t}=a \omega \cos \omega t$ $\Rightarrow \quad \frac{\mathrm{V}}{a \omega}=\cos \omega t$ From eq. (i) and (ii) $$ \begin{gathered} \frac{x^2}{a^2}+\frac{v^2}{a^2 \omega^2}=\sin ^2 \omega t+\cos ^2 \omega t \\ \frac{x^2}{a^2}+\frac{v^2}{a^2 \omega^2}=1 \end{gathered} $$ $\Rightarrow$ This is the equation of ellipse. So the graph between moment and displacement of particle is ellipse.

Hence, the correct answer is option: 'C'.

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