A) \[\Omega \] km and \[{{\tan }^{-1}}(5)\]
B) 10 km and \[{{\tan }^{-1}}\left( \sqrt{5} \right)\]
C) \[\sqrt{50}\] km and \[{{\tan }^{-1}}(5)\]
D) \[\sqrt{5}\]km and \[{{\tan }^{-1}}\left( \sqrt{5} \right)\]
Correct Answer: C
Solution :
Net movement along x-direction \[\Rightarrow \] \[F=ma\] Net movement along y-direction \[=m\frac{{{d}^{2}}x}{d{{t}^{2}}}\] \[=-m{{\omega }^{2}}x\] Net movement from starting point \[\omega =\sqrt{\frac{3kx}{m}}\] \[\Rightarrow \] Angle which resultant makes with the east direction \[T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{3kx}}\] \[=2\pi \sqrt{\frac{m}{3k(sin\omega t)}}\] \[\Rightarrow \] \[T\propto \frac{1}{\sqrt{a}}\]You need to login to perform this action.
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