Two particles are executing identical simple harmonic motions described by the equations `x_1=acos(omegat+((pi)/(6)))` and `x_2=acos(omegat+(pi)/(3)`). The minimum interval of time between the particles crossing the respective mean positions is `(pi)/(2omega)`

Options

A. `(pi)/(2omega)`
B. `(pi)/(3omega)`
C. `(pi)/(4omega)`
D. `(pi)/(6omega)`

Answer

D

Explanation

Equations are `x_1=acos(omegat+(pi)/(6))`
and `x_2=acos(omegat+(pi)/(3))`
The first will pass through the mean position when `x_1=0`
i.e, for instant `t` for which `(omegat+(pi)/(6))=(npi)/(2)`, where `n` is an integer.
The smallest value of `t` is `n=1`,`omegat_1=((pi)/(2))-((pi)/(3))=(pi)/(3)`.
The second will pass through the mean position when
`x_2=0` i.e, for instant `t` for which `(omegat+(pi)/(3))=(mpi)/(2)`
Where `m` is an integer.
The smallest value of `t` is `m=1`,`=((pi)/(2))-((pi)/(3))=(pi)/(6)`
The smallest interval between the instants `x_1=0` and `x_2=0` is therefore,
`omega(t_1:t_2)=((pi)/(3)-(pi)/(6))=(pi)/(6)impliest_1:t_2=(pi)/(6omega)`