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Question
The PE of a particle of mass 0.1 kg moving along the x-axis is given by U = 5x(x – 4) joule, where x is in metre, it can be concluded that
The PE of a particle of mass 0.1 kg moving along the x-axis is given by U = 5x(x – 4) joule, where x is in metre, it can be concluded that
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Solution
The correct option is B The speed of particle is maximum at x = 2m
F=−dUdx=−[5×2x−20]=−10x+20a=Fm=−100x+200=−100(x−2)=−100 X
Where, X=x-2
As a ∝x, so the particle performs SHM
At mean position, X = 0, i.e. x = 2m, so speed of the particle is maximum at x = 2m.
Time period of oscillation is given by, ω2=100
2πT=10⇒T=π5s
F=−dUdx=−[5×2x−20]=−10x+20a=Fm=−100x+200=−100(x−2)=−100 X
Where, X=x-2
As a ∝x, so the particle performs SHM
At mean position, X = 0, i.e. x = 2m, so speed of the particle is maximum at x = 2m.
Time period of oscillation is given by, ω2=100
2πT=10⇒T=π5s
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