A) \[\frac{1}{4}E\]
B) \[\frac{1}{2}E\]
C) \[\frac{2}{3}E\]
D) \[\frac{1}{8}E3\] (where E is the total energy)
Correct Answer: A
Solution :
Potential energy of a simple harmonic oscillator \[U=\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}\] Kinetic energy of a simple harmonic oscillator \[K=\frac{1}{2}m{{\omega }^{2}}({{A}^{2}}-{{y}^{2}})\] Here, y = displacement from mean position A = maximum displacement from mean position (or amplitude) Total energy, \[E=U+K\] \[=\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}+\frac{1}{2}m{{\omega }^{2}}({{A}^{2}}-{{y}^{2}})\] \[=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}\] When the particle is halfway to its end point ie, at half of its amplitude, then \[y=\frac{A}{2}\] Hence, potential energy \[U=\frac{1}{2}m{{\omega }^{2}}{{\left( \frac{A}{2} \right)}^{2}}\] \[=\frac{1}{4}\left( \frac{1}{2}m{{\omega }^{2}}{{A}^{2}} \right)\] \[U=\frac{E}{4}\]
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