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Q.

A particle free to move along the x-axis has potential energy given by
U(x)=k1-exp-x2 for -x+, where k is a constant of appropriate dimensions. Then

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a

at points away from the origin, the particle is in unstable equilibrium

b

for any finite nonzero value of x, there is a force directed away from the origin

c

for small displacements from x=0, the motion is simple harmonic.

d

if its total mechanical energy is k/2, it has its minimum kinetic energy at the origin

answer is D.

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Detailed Solution

Figure shows the plot of U(x) versus x.
At x=0, potential energy U(0)=k[1-exp(0)]=k(1-1)=0 and it has a maximum value =k at x=± since
U(±)=k1-exp(-±)2=k(1-0)=k
Since the total mechanical energy has a constant value =(k/2), the kinetic energy will be maximum at x=0 and minimum at x=±. At x=0
dUdxx=0=2kxexp-x2atx=0=0

Question Image


Hence the particle is in stable equilibrium at x=0 (origin) and would oscillate about x=0 (for small displacements) simple harmonically.

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