A point mass is subjected to two simultaneous sinusoidal displacement in x-direction, `x_(1)(t)=Asinomegat` and `x_(2)(t)=Asin(omegat+2pi//3)`. Adding a third sinusoidal displacement `x_(3)(t)=Bsin(omegat+phi)` brings the mass to a complete rest. The values of B and `phi` are :

Options

A. `sqrt2A`,`(3pi)/(4)`
B. A,`(4pi)/(3)`
C. `sqrt3A,(5pi)/(6)`
D. A,`(pi)/(3)`

Explanation

To solve the problem, we need to analyze the three sinusoidal displacements acting on the point mass and determine the values of \( B \) and \( \phi \) such that the mass comes to rest.

### Step-by-Step Solution:

1. **Identify the Given Displacements:**
The two initial displacements are given as:
\[ x_1(t) = A \sin(\omega t) \]
\[ x_2(t) = A \sin\left(\omega t + \frac{2\pi}{3}\right) \]

2. **Calculate the Resultant of the First Two Displacements:**
To find the resultant of \( x_1(t) \) and \( x_2(t) \), we can use the principle of vector addition. The angle between \( x_1(t) \) and \( x_2(t) \) is \( 120^\circ \) (or \( \frac{2\pi}{3} \) radians). The magnitude of the resultant \( R \) can be calculated using the formula:
\[ R = \sqrt{A^2 + A^2 + 2A \cdot A \cdot \cos(120^\circ)} \]
Since \( \cos(120^\circ) = -\frac{1}{2} \):
\[ R = \sqrt{A^2 + A^2 - A^2} = \sqrt{A^2} = A \]

3. **Determine the Direction of the Resultant:**
The resultant vector \( R \) will be at an angle of \( 60^\circ \) (or \( \frac{\pi}{3} \) radians) from \( x_1(t) \) in the direction of \( x_2(t) \). Therefore, the angle corresponding to this resultant can be expressed as:
\[ \text{Direction of } R = \frac{\pi}{3} \text{ radians} \]

4. **Introduce the Third Displacement:**
The third displacement is given as:
\[ x_3(t) = B \sin(\omega t + \phi) \]
To bring the mass to rest, \( x_3(t) \) must be equal in magnitude but opposite in direction to the resultant \( R \).

5. **Set the Magnitude and Direction of \( x_3(t) \):**
Since the resultant has a magnitude of \( A \), we set:
\[ B = A \]
The direction of \( x_3(t) \) must be opposite to \( R \). Therefore, if \( R \) is at \( \frac{\pi}{3} \), then \( x_3(t) \) should be at:
\[ \phi = \frac{\pi}{3} + \pi = \frac{4\pi}{3} \]

6. **Final Values:**
Thus, we find:
\[ B = A \]
\[ \phi = \frac{4\pi}{3} \]

### Conclusion:
The values of \( B \) and \( \phi \) are:
\[ B = A, \quad \phi = \frac{4\pi}{3} \]