Answer:
As is
known, the particle will turn back when whole of its energy is converted into
potential energy, i.e., \[V(x)=\frac{1}{2}k{{x}^{2}}=1\left(
joule \right)\]
\[\frac{1}{2}\times \frac{1}{2}{{x}^{2}}=1\] or \[{{x}^{2}}=4\]
\[x=\pm 2m\]
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