A particle free to move along the $x$ -axis has potential energy given as $U\left( x \right) = k\left[ {1 - \exp \left( { - {x^2}} \right)} \right]\,\,\,\,\,\,\left( {{\rm{for}} - \infty \le + \infty } \right)$ Where k is a positive constant of appropriate dimensions. Then

At points away from origin, the particle is in equilibrium

For any finite non-zero value of $x$, there is a force directed away from the origin

Its total mechanical energy is $K/2$ and it is equal to its kinetic energy at origin

NEET UG PHYSICS MCQ: OSCILLATIONS

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Analyzing +4 | -1

A particle free to move along the

x

-axis has potential energy given as

U\left( x \right) = k\left[ {1 - \exp \left( { - {x^2}} \right)} \right]\,\,\,\,\,\,\left( {\mathrm{for} - \infty \le + \infty } \right)

Where k is a positive constant of appropriate dimensions. Then

View answer and explanation

Correct option: D

x = 0

, the motion of the particle is simple harmonic

Option D (At

x = 0

, the motion of the particle is simple harmonic) is correct because the potential energy function resembles that of SHM near the origin.

Practice All Chapter MCQs →

{f_1}

{f_2}

F=-k x^3

(k=

)

f

2 f

2 f.

y_1=10 \sin \left(10 \pi t+\frac{\pi}{3}\right)

y_2=12 \sin \left(8 \pi t+\frac{\pi}{4}\right)

t=0.5 \mathrm{~s}