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jimmyu wazir Grade:
        

 










  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

7 years ago

Answers : (1)

Badiuddin askIITians.ismu Expert
147 Points
										

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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7 years ago
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