A) \[\pm \frac{a}{2}\]
B) \[\pm a\]
C) \[\pm 2a\]
D) \[\pm 1\]
Correct Answer: B
Solution :
Key Idea: Potential energy is maximum at extreme positions and kinetic energy is maximum at mean position. Expression of kinetic energy is \[K=\frac{1}{2}\,k\,({{a}^{2}}-{{y}^{2}})...(i)\] Expression of potential energy is \[U=\frac{1}{2}\,k\,{{y}^{2}}....(ii)\] where \[k=m{{\omega }^{2}}\] We observe that at mean position (y = 0) kinetic energy is maximum \[\left( \frac{1}{2}k{{a}^{2}} \right)\]and potential energy is minimum (zero). Also at extreme positions \[(y=\pm a)\] kinetic energy is zero and potential energy is maximum \[\left( \frac{1}{2}\,k{{a}^{2}} \right)\]. Thus, displacement between positions of maximum potential energy and maximum kinetic energy is ± a. Note: Kinetic energy is zero at extreme positions but potential energy at mean position in need not be zero. It is minim at mean position.
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