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the potential energy of theparticle moving along the x-axis is given by U(X)=8x^2+2x^4 where U is in the joule and x is in m.if total mechinal energy is 9j,then limits of motion are

 










  the potential energy of theparticle moving along the x-axis is given by U(X)=8x^2+2x^4 where U is in the joule and x is in m.if total mechinal energy is 9j,then limits of motion are

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1 Answers

Badiuddin askIITians.ismu Expert
147 Points
11 years ago

Dear jimmyu

mechanical energy of ay particle is equal to the sum of kinetic energy and potential energy

so                   1/2mV2 + U =9

 or                   1/2mV2 + 8x2+2x4  =9

 or                    1/2mV2 = 9 -  8x2-2x4 

LHS is a positive value so RHs should also be positive value

               9 -  8x2-2x   ≥0

  or     -√17 -2√2        ≤ x2 ≤   √17- 2√2

 but  0 ≤ x2

  so                             0≤  x2 ≤   √17- 2√2

     now   x2 ≤   √17- 2√2

             x2 -(  √17- 2√2)  ≤0

             x2 -{√(√17- 2√2)}2  ≤0

                      -√(√17- 2√2)   ≤ x   ≤       √(√17- 2√2)

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Badiuddin











 

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