| 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: [JEE Online 15-04-2018] |
| [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], [D]
B) [B], [C]
C) [A], [C], [D]
D) [A], [D]
Correct Answer: C
Solution :
| [c] For mean, |
| \[\Rightarrow \frac{-1}{10}+\frac{1}{v}=-\frac{1}{5}\] |
| \[\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{5}\] |
| \[\Rightarrow \frac{1}{v}=\frac{1}{5}\left[ \frac{1}{2}-1 \right]\] |
| \[\Rightarrow \frac{1}{v}=\frac{-1}{10}\] |
| \[\Rightarrow v=-10cm\] |
| As image copies the time period of object [a] is right as well. It will be periodic motion. |
| For one extreme |
| \[\Rightarrow \frac{-1}{8}+\frac{1}{v}=-\frac{1}{5}\] |
| \[\Rightarrow \frac{1}{v}=\frac{1}{8}-\frac{1}{5}\] |
| \[\Rightarrow \frac{1}{v}=-\frac{3}{40}\] |
| Right arrow\[=\frac{-40}{3}cm\] |
| 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 |
| \[\Rightarrow \frac{40}{3}-\frac{60}{7}\] |
| \[\Rightarrow 10\left[ \frac{4}{3}-\frac{6}{7} \right]\] |
| \[10\times \frac{10}{21}\] |
| \[\Rightarrow \frac{100}{21}cm\] [d] is right |
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