A block of mass m attached to a massless spring is performing oscillatory motion of amplitude ' A ' on a frictionless horizontal plane.

(1) Potential energy of spring $=\frac{1}{2} k x^{2}$

Here, $x=$ distance of block from mean position, $k=$ spring constant

At mean position, potential energy $=\frac{1}{2} k A^{2}$

At equilibrium position, half of the mass of block breaks off, so its potential energy becomes half.

Remaining energy $=\frac{1}{2}\left(\frac{1}{2} k A^{2}\right)=\frac{1}{2} k A^{\prime 2}$

Here, $A^{\prime}=$ New distance of block from mean position

$\Rightarrow A^{\prime}=\frac{A}{\sqrt{2}}$