Question

A particle is oscillating on the X-axis with an amplitude $2cm$ about the point $x_0=10cm$, 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 $x=10cm$.
(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 $=5cm;$ amplitude of oscillation $=2cm;$ position $x_0=10cm$

Therefore, $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$
$\Rightarrow \frac{-1}{5}=\frac{-1}{10}+\frac{1}{v}$
$\Rightarrow \frac{1}{v}=\frac{-1}{10}$
$\Rightarrow v=-10cm$
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 $=\frac{100}{21}cm$.