A) 8 cm/s
B) 12 cm/s
C) 16 cm/s
D) 24 cm/s
Correct Answer: A
Solution :
\[A={{\omega }^{2}}y\] Þ \[\omega =\sqrt{A/y}=\sqrt{\frac{8}{2}}=2\,rad/sec\] Now \[{{v}_{\max }}=a\omega =6\times 2=12\,cm/sec\]
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