A point mass is subjected to two simultaneous sinusoidal displacements in `x - direction, x_1 (t) = A sin (omega)t and x_2 (t) = A sin ((omega t + (2 pi)/(3))`. Adding a third sinusoidal displacement `x _3 (t) = B sin (omega t + phi)` brings the mas to a complete rest. The values of (B) and (phi) are.

Options

A. `sqrt(2)A, (3pi)/(4)`
B. `A, (4pi)/(3)`
C. `sqrt(3)A, (5pi)/(6)`
D. `A, (pi)/(3)`

Answer

B

Explanation

Given, `x_(1)=Asin omegat, x_(2)=A sin (omegat+(2pi)/(3))`
`:. x_(1)+x_(2)=A sin omegat + A sin (omegat+(2pi)/(3))`
`=Asin omegat +A[sin omegat cos ((2pi)/(3))+cos omegat sin ((2pi)/(3))]`
`=A sin omegat +A[sin omegat xx(-(1)/(2))+cosomegat xx((sqrt(3))/(2))]`
`=A[(1)/(2)sin omegat +((sqrt(3))/(2))cosomegat ]`
`=A[cos ((pi)/(3))sin omegat+sin ((pi)/(3))cos omegat]`
`=`A `sin (omegat+(pi)/(3))`
As per question, `x_(1)+x_(2)+x_(3)=0`
`:. x_(3)=-(x_91)+x_(2))=-Asin(omegat+(4pi)/(3))`
`=A sin [omegat +(pi)/(3)+pi]=A sin (omegat+(4pi)/(3))`
or B `sin (omegat+phi)=A sin (omegat+(4pi)/(3))`
`:. B=A, phi=(4pi)/(3)`