Saved Bookmarks
| 1. |
A particle is moving in SHM in a straight line. When the distance of the particle from equilibrium position has values `x_(1)` and `x_(2)` , the corresponding values of velocities are `u_(1)` and `u_(2)`. Show that time period of vibration is `T=2pi[(x_(2)^(2)-x_(1)^(2))/(u_(1)^(2)-u_(2)^(2))]^(1//2)`A. `2pisqrt((x_(2)^(2)-x_(1)^(2))/(upsilon_(1)^(2)-upsilon_(2)^(2)))`B. `2pisqrt((upsilon_(1)^(2)+upsilon_(2)^(2))/(x_(1)^(2)+x_(2)^(2)))`C. `2pisqrt((upsilon_(1)^(2)-upsilon_(2)^(2))/(x_(1)^(2)-x_(2)^(2)))`D. `2pisqrt((x_(1)^(2)+x_(2)^(2))/(upsilon_(1)^(2)+upsilon_(2)^(2)))` |
|
Answer» Correct Answer - A Foa a particle undergoing SHM `upsilon=omega sqrt(A^(2)-x^(2))` `:. upsilon_(1)=omegasqrt(A^(2)-x_(1)^(2)) and upsilon_(2)=omegasqrt(A^(2)-x_(2)^(2))` Squaring we get, `upsilon_(1)^(2)=omega^(2)(A^(2)-x_(1)^(2))` and `upsilon_(2)^(2)=omega^(2)(A^(2)-x_(2)^(2))` Subtracting we get, `upsilon_(1)^(2)-upsilon_(2)^(2)=omega^(2)(x_(2)^(2)-x_(1)^(2))` or `omega=sqrt((upsilon_(1)^(2)-upsilon_(2)^(2))/(x_(2)^(2)-x_(1)^(2))) or (2pi)/(T)=sqrt((upsilon_(1)^(2)-upsilon_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))` `T=2pisqrt((x_(2)^(2)-x_(1)^(2))/(upsilon_(1)^(2)-upsilon_(2)^(2)))` |
|