Question

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

• 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)
$\because F=\frac{dU}{dx}=(x^3-x)$
$\therefore x(x^2-1)=0\Rightarrow x=0,or\pm 1$
$\frac{d^{2}U}{d{x}^2}=3x^2-1$ at $\frac{d^{2}U}{dx}=+ve$ i.e., at $x=\pm 1$
P.E. is minimum when kinetic energy is maximum
$MinU=-\frac{1}{4}$
$K_{max}+U_{min}\Rightarrow 2$
$K_{max}=\frac{9}{4}$
$\therefore \frac{1}{2}m{V}_{max}^2=\frac{9}{4}$
$\because m=1kg$ (given) $\Rightarrow V_{max}=\frac{3}{\sqrt{2}}m/s$