The total energy of an oscillator is the sum of kinetic energy and potential energy at any instant.Total energy = Kinetic energy + Potential energy$E = \frac{1}{2} m u^{2} + \frac{1}{2} k x^{2}$The particle turn back at the instant, when its velocity becomes zero, $u = 0$,Therefore $E = \frac{1}{2} m u^{2} + \frac{1}{2} k x^{2} = 0 + \frac{1}{2} k x^{2}$∴ $1 = \frac{1}{2} \times \frac{1}{2} x^{2}$ $(\because E = 1 \text{ J}, k = 1/2 \text{ N/m})$$1 = \frac{1}{4} x^{2}$$x^{2} = 4, x = \pm 2$Thus, the particle of total energy 1 J moving under this potential, must turn back at $x = \pm 2$ m.