Question

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

• A. zero
• B. $4\text{cm}$
• C. $6\text{cm}$
• D. $1.5\text{cm}$

Answer 1

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.


Here,
$O$ is the position of the object,
$M^{\prime }$ is the initial position of mirror,
$M$ is the final position of mirror,
$I$ is the initial position of image,
$I^{\prime }$ is the final position of image.

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

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 $6\text{cm}$.