For example , at the time `t=0` , the particle is at its mean position, then at which times, the particle will have potential energy `50%` of its total energy within `t=0` to `t=T` ?
Hint `:` When PE is `50%` of total energy `(1)/( 2) kx^(2) = (1)/(2) . (1)/(2) kA^(2)`

Explanation

To solve the problem of when the potential energy (PE) of a particle in simple harmonic motion is 50% of its total energy (TE), we can follow these steps:

### Step 1: Understand the Energy Relationships
In simple harmonic motion, the potential energy (PE) and total energy (TE) are given by:
- Total Energy (TE) = \(\frac{1}{2} k A^2\)
- Potential Energy (PE) = \(\frac{1}{2} k x^2\)

Where:
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