A particle executes S.H.M. 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

Options

A. `T_1 lt T_2`
B. `T_1 gt T_2`
C. `T_1 = T_2`
D. `T_1 = 2T_2`

Answer

A

Explanation

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)`.