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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 ?

```
4 years ago

Vikas TU
14149 Points
```							At eqm. position the net Force would be zero.As, F = – dU/dx => -(a -2bx) = 0a = 2bxx = a/2b would be the eqm. position.
```
4 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions