A) \[\frac{4\pi }{3}\]
B) \[\frac{3\,}{8}\pi \]
C) \[\frac{7\,}{3}\pi \]
D) \[\frac{8\,\pi }{3}\]
Correct Answer: D
Solution :
\[v=\omega \sqrt{{{A}^{2}}-{{x}^{2}}}\] \[a={{\omega }^{2}}x\] \[v=a\] (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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