An object of mass 0.2 kg executes simple harmonic oscillations along the $X$ - axis with frequency of…

The total energy for simple harmonic motion is conserved and equals the sum of kinetic and potential energy at any point. Given
$KE = 1\ \text{J}$ and $PE = 0.6\ \text{J}$,

$E = KE + PE = 1 + 0.6 = 1.6\ \text{J}$.

The angular frequency is derived from the frequency
$f = \frac{25}{\pi}\ \text{Hz}$ as
$\omega = 2 \pi f = 2\pi \cdot \frac{25}{\pi} = 50\ \text{rad/s}$.

The spring constant is
$k = m\omega^2 = 0.2 \times 50^2 = 0.2 \times 2500 = 500\ \text{N/m}$.

Using the energy formula in terms of amplitude,
$E = \frac{1}{2}kA^2$,

which rearranges to
$A = \sqrt{\frac{2E}{k}} = \sqrt{\frac{2 \times 1.6}{500}} = \sqrt{\frac{3.2}{500}} = \sqrt{0.0064} = 0.08\ \text{m}$.

The amplitude is 0.08 m