A) \[\frac{1}{9}\]
B) 2
C) 1
D) None of these
Correct Answer: D
Solution :
| [d] The maximum kinetic energy of the particle is \[\frac{1}{2}m({{A}^{2}}{{\omega }^{2}}).\] |
| The potential energy of the particle at any time t is\[\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\]Using energy conservation |
| \[\Rightarrow \]\[\frac{KE}{PE}=\frac{K{{E}_{\max }}}{PE}-1\] |
| \[\Rightarrow \]\[\frac{KE}{PE}=\frac{\frac{1}{2}m{{A}^{2}}{{\omega }^{2}}}{\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}}-1\] |
| \[=\frac{{{A}^{2}}}{{{A}^{2}}{{\sin }^{2}}\frac{\pi }{90}\times 210}-1\] |
| \[=\frac{1}{{{\left[ \sin \left( 2\pi +\frac{\pi }{3} \right) \right]}^{2}}}-1=\frac{1}{3}\] |
| * None of the given options is correct. |
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