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(C) Let the position of the object be at $x = -d$ (where $d > 0$).Let the mirror perform $SHM$ about the origin $x = 0$ with amplitude $A_m = 2 \ cm$.

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Let the position of the object be at $x = -d$ (where $d > 0$).Let the mirror perform $SHM$ about the origin $x = 0$ with amplitude $A_m = 2 \ cm$.The position of the mirror at any time $t$ is $x_m(t) = A_m \sin(\omega t)$.The position of the image $x_i$ is given by the property that the mirror is always the midpoint between the object and the image.Thus,$x_m = \frac{x_o + x_i}{2}$,where $x_o$ is the constant position of the object.$x_i = 2x_m - x_o = 2(A_m \sin(\omega t)) - x_o$.The image position $x_i(t) = 2A_m \sin(\omega t) - x_o$.The amplitude of the image's $SHM$ is the coefficient of the oscillating term,which is $2A_m$.Given $A_m = 2 \ cm$,the amplitude of the image is $2 	imes 2 \ cm = 4 \ cm$.

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