A) \[\frac{1}{4}F\]
B) \[\frac{1}{2}F\]
C) \[\frac{2}{3}F\]
D) \[\frac{1}{8}F\] (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 (or amplitude) from mean position Total energy is 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 i.e., 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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