InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Two masses `m_(1)and M_(2)` are suspended together by a massless spring of constant k. When the masses are in equilibrium,`m_(1)`is removed without disturbing the system. Then the angular frequency of oscillation of`m_(2)` is |
|
Answer» Let `l_(1)` and `l_(2)` be the extension produced by masses `m_(1)` and `m_(2)` respectively. Then, `(m_(1) +m_(2))g=k(l_(1)+l_(2))…(i)` After `m_(1)` is removed , `m_(2)g=kl_(2)`…(ii) Eqs. (i) -(ii) gives `m_(1)g=kl_(1)` or `l_(1)=(m_(1)g)/(k)` Now, angular frequency of `m_(2), omega=(2pi)/(T)=2pisqrt((k)/(m_(2)))` |
|