Question

The potential energy of a particle moving along the x axis is given by

Answer 1

limits of motion are -0.9568 to +0.9568

The potential energy of a particle moving along the x axis is given by U(x)= 8x² + 2x⁴.. if the total mechanical energy is 9.0 joule,the limits of motion are...

total mechanical energy is sum of potential energy and kinetic energy of the body.

we get limit where kinetic energy of particle becomes zero.

here when kinetic energy of particle equals zero then, potential energy becomes total mechanical energy.

so, 9 Joule = 8x² + 2x⁴

⇒9 = 2x⁴ + 8x²

⇒2x⁴ + 8x² - 9 = 0

let p = x²

then, 2p² + 8p - 9 = 0

p = {-8 ± √(64 + 72)}/4 = -4.915 , 0.91548

but p = x²

so, p ≠ -4.915

then, 0.91548 = p = x²

x² = 0.91548

x = ±√(0.91548) = ± 0.9568

hence, limits of motion are -0.9568 to +0.9568

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Answer 2

$\huge\bold\purple{Answer:-}$

Potential energy of the system is given as

U = 8x^2 + 2x^4

its mechanical energy is also given as

E = 9J

now when whole mechanical energy will convert to Potential energy then it will be the limit of motion

9 = 8x^2 + 2x^4

2x^4 + 8x^2 - 9 = 0

by solving above equation we have

x^2 = 0.91

-0.95 < x < + 0.95

so it will move from x = -0.95 m to x = + 0.95 m