A) \[\frac{\pi }{5}\]sec
B) \[2\pi \]sec
C) \[20\pi \]sec
D) \[5\pi \]sec
Correct Answer: A
Solution :
At mean position, the kinetic energy is maximum. Hence \[\frac{1}{2}m{{a}^{2}}{{\omega }^{2}}=16\] On putting the values we get \[\omega =10\ \Rightarrow \ T=\frac{2\pi }{\omega }=\frac{\pi }{5}\sec \]
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