Explanation

(a) Potential energy of particle in SHM
$U = \frac{1}{2}m\omega^{2}x^{2}$
or $U = \frac{1}{2}m(2\pi f)^{2}x^{2}$
or $U = 2\pi^{2}mf^{2}x^{2}$ …(i)
Kinetic energy of particle in SHM
$K = \frac{1}{2}m\omega^{2}(A^{2} - x^{2})$
or $K = 2\pi^{2}mf^{2}(A^{2} - x^{2})$ …(ii)
Hence, total energy
$E = K + U = 2\pi^{2}mf^{2}x^{2} + 2\pi^{2}mf^{2}(A^{2} - x^{2})$
$= \ 2\pi^{2}mf^{2}A^{2} = \frac{2\pi^{2}mA^{2}}{T^{2}}$ $\left( \because\ \ T = \frac{1}{f} \right)$
Thus, it is obvious that total energy of particle executing simple harmonic motion depends on amplitude ($A$) and period ($T$).