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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