Question

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

• A. $\frac{3}{\sqrt{2}}\text{m/s}$
• B. $\sqrt{2}\text{m/s}$
• C. $\frac{1}{\sqrt{2}}\text{m/s}$
• D. $2\text{m/s}$

Answer 1

The correct option is A $\frac{3}{\sqrt{2}}\text{m/s}$

The speed of the particle is maximum when its kinetic energy is maximum and from energy conservation law, the potential energy must be minimum at this point. For minimum value of potential energy
$\frac{dV}{dx}=0$
$\frac{d}{dx}[\frac{x^4}{4}-\frac{x^2}{2}]=0$
$\Rightarrow x^3-x=0\Rightarrow x(x^2-1)=0$
$\Rightarrow x=0$ and $x=\pm 1$
Now $V_{\text{at}(x=0)}=0$ and $V_{(x=\pm 1)}=(\frac{1}{4}-\frac{1}{2})=-\frac{1}{4}\text{J}$
Thus, the minimum possible value of potential energy is $-\frac{1}{4}\text{J}$
Since total energy = Kinetic energy + Potential energy
$\therefore$ KE = TE - PE $=2-(-\frac{1}{4})=\frac{9}{4}\text{J}$
$\therefore \frac{1}{2}m{v}^2=\frac{9}{4}\Rightarrow v=\sqrt{\frac{9}{2m}}=\sqrt{\frac{9}{2\times 1}}=\frac{3}{\sqrt{2}}\text{m/s}$
Hence, the correct answer is option (a).