Question

A particle is oscillating on the X-axis with an amplitude $2 \,cm$ about the point $x_0=10 \, cm$, with a frequency $\omega$. A concave mirror of focal length 5 cm is placed at the origin (see figure).
Identify the correct statements.
(A) The image executes periodic motion.
(B) The image executes non-periodic motion.
(C) The turning points of the image are asymmetric w.r.t. the image of the point at $x = 10 \, cm$.
(D) The distance between the turning points of the oscillation of the image is $\frac{100}{21} $ cm.

• A. (A), (D)
• B. (A), (C), (D)
• C. (B), (D)
• D. (B), (C)

Answer 1

Answer: (B)

Given, focal length $=5 cm ;$ amplitude of oscillation $=2\, cm ;$ position $x_{0}=10\, cm$

Therefore, $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$
$\Rightarrow \frac{-1}{5}=\frac{-1}{10}+\frac{1}{v}$
$\Rightarrow \frac{1}{v}=\frac{-1}{10}$
$\Rightarrow v=-10\, cm$
Thus, image executes a periodic motion, so (A) is correct.
Now since the particle is oscillating,
For right extreme,
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
$\Rightarrow \frac{-1}{5}=\frac{1}{v}-\frac{1}{8}$
$\Rightarrow \frac{1}{v}=\frac{-3}{40}$
$\Rightarrow v_{ R }=\frac{40}{3} cm$
For left extreme,
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}=\frac{-1}{5}=\frac{1}{v}-\frac{1}{12}$
$\Rightarrow \frac{1}{v}=\frac{-7}{60}$
$\Rightarrow v_{ L }=-\frac{60}{7}\, cm$
Thus, the points are symmetric about $x_{0}$, option (C) is correct.
Now amplitude of image $=v_{ L }-v_{ R }=\frac{-60}{7}-\left(\frac{-40}{3}\right)=\frac{40}{3}-\frac{60}{7}=\frac{100}{21}\, cm$
Therefore, option (D) is correct distance between turning points of image $=\frac{100}{21}\, cm$.