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.

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

Any SHM is given by the equation `x=sinomegat`, where x is the displacement of the body any instant t. A is the amplitude and `omega` is the angular frequency.
when `x=0,omegat_(1)=0`
`thereforet_(1)=0`
when `x=A//2,omegat_(2)=pi//6,t_(2)=pi//6omega`
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)/(6omega)=T_(1)`
time taken from A/2 to A will be
`t_(3)-t_(2)=(pi)/(2omega)-(pi)/(6omega)=(2pi)/(6omega)=(pi)/(3omega)=T_(2)`
Hence `T_(2)gtT_(1)`