Saved Bookmarks
| 1. |
A particle executes simple harmonic motion between `x = -A and x = + A`. The time taken for it to go from `0 to A//2 is T_1 and to go from A//2 to (A) is (T_2)`. Then.A. `T_1 lt T_2`B. `T_1 gt T_2`C. `T_1 = T_2`D. `T_1 = 2T_2` |
|
Answer» Correct Answer - A Method 1 : The velocity of a body executing `SHM` is maximum at its centre and decreases as the body proceeds to the extremes. Therefore, if the time taken for the body to go from `O` to `A//2` is `T_(1)` and to go `A` is `T_(2)`, then obviously `T_(1) lt T_(2)`. Method 2 : Quantitative. Any `SHM` is given by the equation `x = sin omega t`. where `x` is the displacement of the body at any instant `t.a` is the amplitude and `omega` is the angular frequency. When `x =0, omega t_(1) = 0` `:. t_(1) = 0` `{:(When,x=a//2,omegat_(2)=pi//6,t_(2)=pi//6 omega,),(When,x=a,omegat_(3)=pi//2,t_(3)=pi//2omega,):}` Time taken from `O` to `A//2` will be `t_(2) - t_(1) = (pi)/(6 omega) = T_(1)` Time taken from `A//2` to `A` will be `t_(3)-t_(2)=(pi)/(2 omega)-(pi)/(6 omega)=(2pi)/(6 omega)=(pi)/(3 omega) = T_(2)` Hence `T_(2) gt T_(1)`. |
|