A) 5 J
B) 10 J
C) 15 J
D) 20 J
Correct Answer: B
Solution :
Potential energy of particle is P.E. \[=\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}=2.5\] \[\Rightarrow \] \[\frac{1}{2}m\,{{\omega }^{2}}{{\left( \frac{a}{2} \right)}^{2}}=2.5\] \[\left( \because \,y=\frac{a}{2} \right)\] \[\Rightarrow \] \[\left( \frac{1}{2}m\,{{\omega }^{2}} \right).\frac{{{a}^{2}}}{4}=2.5\] ?(1) Hence, total energy is \[E=\frac{1}{2}m\,{{\omega }^{2}}{{a}^{2}}\] [From Eq. (1)] \[=2.5\times 4\] \[=10\,J\]
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