QJEEM18Phefab: JEE Main 2018 Physics - Class 12 JEEM18Phefab Solution: A particle is oscillating on the -axis with an amplitude about…

Given: Focal length $f = 5$ cm, Mean position $x_0 = 10$ cm, Amplitude $A = 2$ cm.

The object oscillates between $u = 10 - 2 = 8$ cm and $u = 10 + 2 = 12$ cm.

Using mirror formula: $v = \frac{uf}{u - f}$

When object is at 8 cm,

the image distance (v₁) = ${{u \times f} \over {u - f}}$ = ${{8 \times 5} \over {8 - 5}}$ = ${{40} \over 3}$ cm

When object is at 12 cm,

the image distance (v₂) = ${{u \times f} \over {u - f}}$ = ${{12 \times 5} \over {12 - 5}}$ = ${{60} \over 7}$ cm

$\therefore\,\,\,\,$ Separation = $\left| {v_1 - v_2} \right|$ = ${{40} \over 3} - {{60} \over 7}$ = ${{100} \over {21}}\,$ cm

For mean position $x = 10$ cm:

v_mean = ${{10 \times 5} \over {10 - 5}}$ = $10$ cm

Since $|v_1 - v_{mean}| \neq |v_2 - v_{mean}|$ , the turning points are asymmetric w.r.t. the image of the mean position.

As the object's motion is periodic, the image's motion is also periodic.

Correct Options: (A), (C), (D)