The plane mirror shown in above figure is performing SHM with amplitude $A=3\text{cm}$. The amplitude of SHM of image wrt ground is

The correct option is C $6\text{cm}$
Let the initial distance between mirror and object $=x\text{cm}$.
Along the normal direction for plane mirror, object distance = image distance.



The amplitude of SHM for image will be, $A^{\prime }=I{I}^{\prime }$
$\Rightarrow I{I}^{\prime }=OI-O{I}^{\prime }$
$\Rightarrow I{I}^{\prime }=x-\left[\right(x-3)-A]=x-\left[\right(x-3)-3]$
$\therefore I{I}^{\prime }=x-x+6=6\text{cm}$

Or

As object is rest, if mirror moves a distance $A$ along its normal from its position, image moves a distance $2A$
Hence amplitude of oscillation wrt ground is $6\text{cm}$