A) \[\sqrt{3}\,cm\]
B) \[\sqrt{5}\,cm\]
C) \[2(\sqrt{3})\,cm\]
D) \[2(\sqrt{5})\,cm\]
Correct Answer: C
Solution :
\[{{v}_{\max }}=a\omega \] Þ \[\omega =\frac{{{v}_{\max }}}{a}=\frac{10}{4}\] Now, \[v=\omega \sqrt{{{a}^{2}}-{{y}^{2}}}\]Þ \[{{v}^{2}}={{\omega }^{2}}({{a}^{2}}-{{y}^{2}})\]Þ \[{{y}^{2}}={{a}^{2}}-\frac{{{v}^{2}}}{{{\omega }^{2}}}\] Þ \[T\propto \sqrt{m}\]\[=2\sqrt{3}\]cm
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