InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
When the potential energy of a particle executing simple harmonic motion is one-fourth of its maximum value during the oscillation, the displacement of the particle from the equilibrium position in terms of its amplitude a isA. `(A)/(4)`B. `(A)/(3)`C. `(A)/(2)`D. `(A)/(sqrt(2))` |
|
Answer» Correct Answer - D Potential energy of particle, `U = (1)/(2)m omega^(2)y^(2)` Potential energy of maximum particle, `E = (1)/(2)m omega^(2)A^(2)` According to given position, the potential energy, `U = (E)/(2)` or `(1)/(2)m omega^(2)y^(2)=(1)/(2)xx(1)/(2)m omega^(2)A^(2)` `rArr" "y^(2)=(A^(2))/(2),y=(A)/(sqrt(2))` |
|