A particle with total mechanical energy, which is small and negat… - Vidyadip

MCQ

A particle with total mechanical energy, which is small and negative. Its under the influence of a one dimensional potential $U(x)=x^{4} / 4-x^{2} / 2 \,J$, where $x$ is (in metre). At time $t=0 \,s$, it is at $x=-0.5 \,m$. Then, at a later time it can be found

A
anywhere on the $X$-axis
B
between $x=-1.0 \,m$ to $x=1.0 \,m$
✓
between $x=-1.0 \,m$ to $x=0.0 \,m$
D
between $x=0.0 \,m$ to $x=1.0 \,m$

✓

Answer

Correct option: C.

between $x=-1.0 \,m$ to $x=0.0 \,m$

c
$(c)$ Given, potential energy of particle is

$U=\frac{x^{4}}{4}-\frac{x^{2}}{2}$

Differentiating with respect to $x$, we get

$\frac{d U}{d x}=\frac{4 x^{3}}{4}-\frac{2 x}{2}$

$=x^{3}-x$

$=x\left(x^{2}-1\right)$

For maxima or minima,

$\frac{d U}{d x} =0$

$\Rightarrow x\left(x^{2}-1\right) =0$

$\Rightarrow x=0 \text { or } x =\pm 1$

Differentiating again, we have

$\frac{d^{2} \tau}{d x^{2}}=3 x^{2}-1$

So, $\quad \frac{d^{2} U}{d x^{2}}=\left\{\begin{array}{c}-1 \text { at } x=0 \\ +2 \text { at } x=\pm 1\end{array}\right.$

So, by second derivative test, potential energy maxima occurs at $x=\pm 1$ and potential energy minima is at $x=0$.

Now, as at $t=0, x=-0.5 \,m$, so the particle is region $x=-1$ to $x=0$. As total encrgy of particle is negative, so particle cannot crosses $x=0$ mark

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