The potential energy function for a particle executing simple harmonic motion is given by `V(x)=(1)/(2)kx^(2)`, where k is the force constant of the oscillatore. For `k=(1)/(2)Nm^(-1)`, show that a particle of total energy 1 joule moving under this potential must turn back when it reaches `x=+-2m.`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Energy Conservation Principle
The total mechanical energy \(E\) of a particle in simple harmonic motion is the sum of its kinetic energy \(K\) and potential energy \(V\):
\[ E = K + V \]
Given that the total energy \(E\) is 1 joule, we can express this as:
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