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The potential energt of a particle of mass 0.1 kg, moving along the x-axis, is given by `U=5x(x-4)J`, where x is in meter. It can be concluded thatA. the particle is acted upon by a constant forceB. the speed of the particle is maximum at `x=2m`C. the particle executes SHMD. the period of oscillation of the particle `((pi)/(5))` s |
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Answer» Correct Answer - B::C::D `U=5x(x-4)=5x^2-20x` `F=-(dU)/(dx)=-10x+20` i.e., force is not contant. KE or speed of the particle will be maximum at the mean position where force becomes zero. `F=0` or `x=2m` Acceleration experienced by the particle is `a=(F)/(m)=(-10x+20)/(0.1)=-(100x-200)` i.e., particle executes SHM. As `omega^2=100` `omega=10` Hence `T=(2pi)/(omega)=(pi)/(5)s` |
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