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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
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 - 8x2-2x4 ≥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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