Question

The potential energy of $1\text{kg}$ particle free to move along the $X-$ axis is given by $U=(\frac{x^4}{4}-\frac{x^2}{2})\text{J}$. The total mechanical energy of the particle is $2\text{J}$. Then maximum speed of the particle is (in $\text{m/s}$)

• A. $\frac{3}{\sqrt{2}}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\sqrt{2}$
• D. $2$

Answer 1

The correct option is A $\frac{3}{\sqrt{2}}$

$U=\frac{x^4}{4}-\frac{x^2}{2}$ (Given)
$F=\frac{dU}{dx}=(x^3-x)$
$\frac{dU}{dx}=0\Rightarrow x(x^2-1)=0$
$\Rightarrow x=0,$ $\pm 1$
$\frac{d^{2}U}{d{x}^2}=3x^2-1$
$\Rightarrow \frac{d^{2}U}{dx}=+ve$ at $x=\pm 1$
(Point of minima)
i.e $U_{min}=-\frac{1}{4}\text{J}$

$K_{\text{max}}+U_{\text{min}}=2\text{J}$
$\because$ P.E. is minimum when kinetic energy is maximum
$\Rightarrow K_{\text{max}}=\frac{9}{4}\text{J}$
i.e $\frac{1}{2}m{V}_{\text{max}}^2=\frac{9}{4}$
$\Rightarrow V_{\text{max}}=\frac{3}{\sqrt{2}}\text{m/s}$
($\because m=1\text{kg}$ (given))