Answer :

Given,

Total energy of the particle, E = 1 J

Force constant, k = 0.5 N m^{-1}

The total energy is equal to the sum of kinetic energy and potential energy.

So, E = PE + K

⇒ E = kx^{2} + mv^{2}

So, 1 = kx^{2} + mv^{2}

When the velocity of the particle is zero i.e., at the turning point, the kinetic energy is zero.

∴ 1 = kx^{2}

⇒ x^{2} = 2/k

⇒ x^{2} = 2/0.5

⇒ x^{2} = 4

⇒ x = �2

Hence, the particle turns back at x= �2.

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