A) \[2\pi \alpha \]
B) \[2\pi \sqrt{\alpha }\]
C) \[\frac{2\pi }{\alpha }\]
D) \[\frac{2\pi }{\sqrt{\alpha }}\]
Correct Answer: D
Solution :
\[\frac{{{d}^{2}}x}{d{{t}^{2}}}\,\,+\,\,{{\omega }^{2}}x=0\] \[\frac{{{d}^{2}}x}{d{{t}^{2}}}\,\,+\,\,\alpha x=0\] \[{{\omega }^{2}}\,\,=\,\,\alpha \] \[\omega \,\,=\,\,\sqrt{\alpha }\,=\,\,\frac{2\pi }{T}\] \[T=\frac{2\pi }{\sqrt{\alpha }}\]
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