A) 6 J
B) 18 J
C) 24 J
D) 36 J
Correct Answer: B
Solution :
So \[a=6cm,\ \omega =100rad/\sec \] \[{{K}_{\max }}=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}=\frac{1}{2}\times 1\times {{(100)}^{2}}\times {{(6\times {{10}^{-2}})}^{2}}=18\ J\]
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