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Question

The potential energy of a particle of mass 0.1 kg, moving along the xaxis, is given by U=5x(x4) J, where x is in meters.
Choose the wrong option.

A
The speed of the particle is maximum at x=2 m.
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B
The particle executes simple harmonic motion.
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C
The period of oscillation of the particle is π5 s.
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D
The particle does not execute simple harmonic motion.
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Solution

The correct option is D The particle does not execute simple harmonic motion.
Given that,
mass of the particle m=0.1 kg
Potential energy is U=5x220x,
For conservative forces,
F=dUdx=(10x20)
F=10(x2)
F=10X (Assume x2=X)
ma=10X (a=acceleration of the particle)
0.1 kg×a=10X
0.1 kg×a=10X
a=100X ...(i)
aX
Particle is performing simple harmonic motion.
Option b is correct.

Comparing equation (i) with simple harmonic motion
a=ω2x=100x
ω=10
Time period of the simple harmonic motion is
T=2πω=2π10
T=π5s
option c is also correct.

At x=2 m,F=0 i.e. particle is at mean position and speed is maximum.
Option a is correct.

Why this question ?

A conservative force at a point is negative of potential energy gradient at that position.

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