Saved Bookmarks
| 1. |
A mass m is hung on an ideal massless spring another equal mass is connected to the other end of the spring the whole system is at rest at t=0 m is released and the system falls freely under gravity assume that natural length of the spring is L_(0)its initial stretched length is L and the acceleration due to gravity is g what is distance between masses as function of time |
|
Answer» `L_(0)+(L-L_(0))cos sqrt((2k)/m)t` `omega=sqrt(k/mu)=sqrt ((2k)/m)` initially PARTICLES are at EXTREME `DISTANCE = L_(0)+(L-L_(0))cos sqrt((2k)/m)t` |
|