the potential energy U(x) of a particle moving along x-axis is given by U(x) = ax-bx2 .find the equillibrium position of the particle ?

Question

Vikas TU · Answer

At eqm. position the net Force would be zero. As, F = – dU/dx => -(a -2bx) = 0 a = 2bx x = a/2b would be the eqm. position.

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the potential energy U(x) of a particle moving along x-axis is given by U(x) = ax-bx2 .find the equillibrium position of the particle ?

the potential energy U(x) of a particle moving along x-axis is given by U(x) = ax-bx2 .find the equillibrium position of the particle ?

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the potential energy U(x) of a particle moving along x-axis is given by U(x) = ax-bx2 .find the equillibrium position of the particle ?

the potential energy U(x) of a particle moving along x-axis is given by U(x) = ax-bx2 .find the equillibrium position of the particle ?

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