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 $5cm$ is placed at the origin (see figure)
Identify the correct statements:
(A) The image executes periodic motion
(B) The image executes no n-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 $\frac{100}{21}$

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

Answer 1

Answer: (B)

For mean ,

$\Rightarrow \frac{-1}{10}+\frac{1}{v}=-\frac{1}{5}$

$\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{5}$

$\Rightarrow \frac{1}{v}=\frac{1}{5}[\frac{1}{2}-1]$

$\Rightarrow \frac{1}{v}=\frac{-1}{10}$

$\Rightarrow v=-10cm$

As image copies the time period of object (A) is right as well. It will be periodic motion.

For one extreme

$\Rightarrow \frac{-1}{8}+\frac{1}{v}=-\frac{1}{5}$

$\Rightarrow \frac{1}{v}=\frac{1}{8}-\frac{1}{5}$

$\Rightarrow \frac{1}{v}=-\frac{3}{40}$

$Rightarrowv=\frac{-40}{3}cm$

For other extreme

$\Rightarrow \frac{-1}{12}+\frac{1}{v}=-\frac{1}{5}$

$\Rightarrow \frac{1}{v}=\frac{1}{12}-\frac{1}{5}$

$\Rightarrow \frac{1}{v}=\frac{-7}{60}c{m}^{-1}$

$\Rightarrow v=\frac{-60}{7}cm$

These points are asymmetric about $x_0=10cm$ So,(c) is right.

Amplitude of oscillation of image

$\Rightarrow \frac{40}{3}-\frac{60}{7}$

$\Rightarrow 10[\frac{4}{3}-\frac{6}{7}]$

$10\times \frac{10}{21}$

$\Rightarrow \frac{100}{21}cm$ (D) is right