SATHEE: Simple Harmonic Motion 5 Question 7

Simple Harmonic Motion 5 Question 7

12. A particle free to move along the $X$ -axis has potential energy given by $U (x) = k [1 - \exp (- x^{2})]$ 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

(d) for small displacements from $x = 0$ , the motion is simple harmonic

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Answer:

Correct Answer: 12. (d)

Solution:

$U (x) = k (1 - e^{- x^{2}})$

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

$F = \frac{- d U}{d x} = - (slope of U - x graph) = 0$

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.

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