A) \[\frac{4\pi }{3}\]
B) \[\frac{3\,}{8}\pi \]
C) \[\frac{7\,}{3}\pi \]
D) \[\frac{8\,\pi }{3}\]
Correct Answer: D
Solution :
| [d] \[v=\omega \sqrt{{{A}^{2}}-{{x}^{2}}}\] |
| \[a={{\omega }^{2}}x\] |
| \[v=a\] (according to question |
| \[\left| velocity \right|\text{ = }\left| acceleration \right|\text{ })\] |
| \[~\omega \sqrt{{{A}^{2}}-{{x}^{2}}}\,=\,{{\omega }^{2}}x\] |
| \[~\sqrt{{{A}^{2}}-{{x}^{2}}}\,=\,\omega x\] |
| \[{{A}^{2}}-{{x}^{2}}={{\omega }^{2}}{{x}^{2}}\] |
| \[25-16={{\omega }^{2}}\times 16\] |
| \[9={{\omega }^{2}}\times 16\] |
| \[\omega \,\,=\,\,\sqrt{\frac{9}{16}}\,\,=\,\frac{3}{4}\] |
| \[T\,\,=\,\,\frac{2\pi }{\omega }\,=\,\frac{2\pi }{3}\,\times 4\,\,=\,\,\frac{8\,\pi }{3}\,\sec \] |
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