Question

The velocity vector $v$ and displacement vector $x$ of a particle executing SHM are related as $\frac{v d v}{d x}=-\omega^{2} X$ with the initial condition $V=v_{0}$ at $x=0$. The velocity $v$, when displacement is $X$, is

• A. $v=\sqrt{v_{0}^{2}+\omega^{2} x^{2}}$
• B. $v=\sqrt{v_{0}^{2}-\omega^{2} x^{2}}$
• C. $v=\sqrt[3]{v_{0}^{3}+(1)^{3} x^{3}}$
• D. $v=v_{0}- (\omega^3 x^3e^{x^3})^{1/3}$

Answer 1

Answer: (B)

Given,
$v \frac{d v}{d x}=-\omega^{2} x$
On integrating within the limit
$ \int\limits_{V_{0}}^{V} v d v =\int\limits_{0}^{x}-\omega^{2} x d x$
$\Rightarrow \left[\frac{v^{2}}{2}\right]_{V_{0}}^{v}=-\omega^{2}\left[\frac{x^{2}}{2}\right]_{0}^{x} $
$\Rightarrow v^{2}-v_{0}^{2} =-\omega^{2} x^{2} $
$\Rightarrow v=\sqrt{v_{0}^{2}-\omega^{2} x^{2}}$