A) \[\left( \frac{1}{8} \right)E\]
B) \[\left( \frac{1}{4} \right)E\]
C) \[\left( \frac{1}{2} \right)E\]
D) \[\left( \frac{2}{3} \right)E\]
Correct Answer: B
Solution :
Total \[{{\varepsilon }_{n}}=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}\] \[P.E.=\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}\] \[\frac{P.E}{\text{Total}\,{{\omega }_{n}}}=\frac{{{y}^{2}}}{{{A}^{2}}}={{\left( \frac{1}{2} \right)}^{2}}=\frac{1}{4}.\] Hence, the correction option is [b].
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