Explanation

(a) If a particle executes SHM, its kinetic energy is given by
$KE = \frac{1}{2}m\omega^{2}(A^{2} - x^{2})$
or $KE = \frac{1}{2}\ k(A^{2} - x^{2})$
where $k = m\omega^{2} = constant$
Its potential energy is given by
$KE = \frac{1}{2}m\omega^{2}x^{2} = \frac{1}{2}\ kx^{2}$
Thus, total energy of particle
$E = KE + PE$
$= \ \frac{1}{2}\ k\left( A^{2} - x^{2} \right) + \frac{1}{2}\ kx^{2}$
$= \ \frac{1}{2}\ kA^{2}$
Hence, $PE = \frac{1}{2}\ kx^{2} = \frac{1}{2}\ k\ \left( \frac{A}{2} \right)^{2}$ ($\because\ \ x = \frac{A}{2}$)
$= \ \frac{1}{4}\ \left( \frac{1}{2}\ kA^{2} \right)$
$= \ \frac{1}{4}\ E$
Hence, potential energy is one-fourth of total energy.