A) P.E. is maximum when x = 0.
B) K.E. is maximum when x = 0.
C) T. E. is zero when x = 0.
D) K.E. is maximum when x is maximum.
Correct Answer: B
Solution :
[b] \[K=\frac{1}{2}m{{\omega }^{2}}({{A}^{2}}-{{x}^{2}})\,\therefore \,{{K}_{\max }}=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}},\] at \[x=0\]You need to login to perform this action.
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