InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where, E is the total energy)A. `(1)/(4)E`B. `(1)/(2)E`C. `(2)/(3)E`D. `(1)/(8)E` |
|
Answer» Correct Answer - A Potential energy of a simple harmonic oscillator `U=(1)/(2)momega^(2)y^(2)` Kinetic energy of a simple harmonic oscillator `K=(1)/(2)momega^(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 `=(1)/(2)momega^(2)y^(2)+(1)/(2)momega^(2)(A^(2)-y^(2))` `=(1)/(2)momega^(2)A^(2)` When the particle half way to its end point ie, at half of its amplitude then `y=(A)/(2)` Hence potential energy `U=(1)/(2)momega^(2)((A)/(2))^(2)` `=(1)/(4)((1)/(2)momega^(2)A^(2))` `U=(E)/(4)` |
|