Explanation

(b) The relation for kinetic energy of S.H.M. is given by
$= \frac{1}{2}m\omega^{2}\left( a^{2} - x^{2} \right)$ …(i)
Potential energy is given by
$= \frac{1}{2}m\omega^{2}x^{2}$ …(ii)
Now for the condition of question and from eqs. (i) and (ii)
$\frac{1}{2}m\omega^{2}\left( a^{2} - x^{2} \right) = \frac{1}{3} \times \frac{1}{2}m\omega^{2}x^{2}$
$or\ \frac{4}{6}m\omega^{2}x^{2} = \frac{1}{2}m\omega^{2}a^{2}\ or\ x^{2} = \frac{3}{4}a^{2}$
So$,\ x = \frac{a}{2}\sqrt{3} = 0.866\ a \approx 87\%\ $of amplitude