A) \[\frac{4a}{T}\]
B) \[\frac{2a}{T}\]
C) \[2\pi \sqrt{\frac{a}{T}}\]
D) \[\frac{2\pi a}{T}\]
Correct Answer: D
Solution :
\[{{v}_{\max }}=a\omega =\frac{a\,.\,2\pi }{T}=\frac{2\pi a}{T}\]
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