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(C) The object oscillates between $x_1 = 8\, cm$ and $x_2 = 12\, cm$. Since the motion is periodic,the image will also execute periodic motion. Thus,s

Vedclass · Accepted Answer

$(A), (C), (D)$

Vedclass · Answer

(C) The object oscillates between $x_1 = 8\, cm$ and $x_2 = 12\, cm$. Since the motion is periodic,the image will also execute periodic motion. Thus,statement $(A)$ is correct and $(B)$ is incorrect.Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,where $f = -5\, cm$ (concave mirror).For $u_1 = -8\, cm$:$\frac{1}{v_1} + \frac{1}{-8} = \frac{1}{-5} \implies \frac{1}{v_1} = \frac{1}{8} - \frac{1}{5} = \frac{5-8}{40} = -\frac{3}{40} \implies v_1 = -\frac{40}{3}\, cm$.For $u_2 = -12\, cm$:$\frac{1}{v_2} + \frac{1}{-12} = \frac{1}{-5} \implies \frac{1}{v_2} = \frac{1}{12} - \frac{1}{5} = \frac{5-12}{60} = -\frac{7}{60} \implies v_2 = -\frac{60}{7}\, cm$.The distance between the turning points is $|v_1 - v_2| = |-\frac{40}{3} - (-\frac{60}{7})| = |-\frac{280}{21} + \frac{180}{21}| = |-\frac{100}{21}| = \frac{100}{21}\, cm$. Thus,statement $(D)$ is correct.The image of the center point $x_0 = -10\, cm$ is $v_0 = \frac{f \cdot u_0}{u_0 - f} = \frac{-5 \cdot -10}{-10 - (-5)} = \frac{50}{-5} = -10\, cm$. The midpoint of the image range is $\frac{v_1 + v_2}{2} = \frac{-40/3 - 60/7}{2} = \frac{-280 - 180}{42} = -\frac{460}{42} = -\frac{230}{21} \approx -10.95\, cm$. Since the midpoint of the image range is not equal to the image of the center point,the motion is asymmetric. Thus,statement $(C)$ is correct.Therefore,statements $(A), (C),$ and $(D)$ are correct.

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