Simple Harmonic Motion 5 Question 7
12. A particle free to move along the $X$-axis has potential energy given by $U(x)=k\left[1-\exp \left(-x^{2}\right)\right]$ for $-\infty \leq x \leq+\infty$, where $k$ is a positive constant of appropriate dimensions. Then,
(1999, 2M)
(a) at points away from the origin, the particle is in unstable equilibrium
(b) for any finite non-zero value of $x$, there is a force directed away from the origin
(c) if its total mechanical energy is $k / 2$, it has its minimum kinetic energy at the origin
(d) for small displacements from $x=0$, the motion is simple harmonic
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Answer:
Correct Answer: 12. (d)
Solution:
- $U(x)=k\left(1-e^{-x^{2}}\right)$
It is an exponentially increasing graph of potential energy $(U)$ with $x^{2}$. Therefore, $U$ versus $x$ graph will be as shown.
At origin.
Potential energy $U$ is minimum (therefore, kinetic energy will be maximum) and force acting on the particle is zero because
$$ F=\frac{-d U}{d x}=-(\text { slope of } U-x \text { graph })=0 $$
Therefore, origin is the stable equilibrium position. Hence, particle will oscillate simple harmonically about $x=0$ for small displacements. Therefore, correct option is (d).
(a), (b) and (c) options are wrong due to following reasons.
(a) At equilibrium position $F=\frac{-d U}{d x}=0$ i.e. slope of $U-x$ graph should be zero and from the graph we can see that slope is zero at $x=0$ and $x= \pm \infty$.
Now, among these equilibriums stable equilibrium position is that where $U$ is minimum (Here $x=0$ ). Unstable equilibrium position is that where $U$ is maximum (Here none).
Neutral equilibrium position is that where $U$ is constant (Here $x= \pm \infty$ ).
Therefore, option (a) is wrong.
(b) For any finite non-zero value of $x$, force is directed towards the origin because origin is in stable equilibrium position. Therefore, option (b) is incorrect.
(c) At origin, potential energy is minimum, hence kinetic energy will be maximum. Therefore, option (c) is also wrong.
