Saved Bookmarks
| 1. |
The ratio of kinetic energy to the potential energy of a particle executing SHM at a distance equal to half its amplitude, the distance being measured from its equilibrium position is |
|
Answer» Solution :Potential energy `U= (1)/(2) KX^(2)= (1)/(2)k (A^(2))/(4)` Where `x = (A)/(2)` Kinetic energy `K = (1)/(2) kA^(2)- (1)/(2) k(A^(2))/(4)= (3)/(8) kA^(2)` Hence `(K)/(U)= (3//8)/(1//8)= 3: 1` |
|