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 find the values of \( B \) and \( \phi \) such that the resultant displacement of the point mass is zero when subjected to the three sinusoidal displacements.

### Step-by-Step Solution:

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

2. **Convert to Phasor Representation**:
The first displacement can be represented as a phasor at an angle of \( 0^\circ \) (or \( 0 \) radians):
- \( x_1 \) has an amplitude \( A \) and angle \( 0 \).

The second displacement can be represented as a phasor at an angle of \( \frac{2\pi}{3} \) (or \( 120^\circ \)):
- \( x_2 \) has an amplitude \( A \) and angle \( \frac{2\pi}{3} \).

3. **Calculate the Resultant of \( x_1 \) and \( x_2 \)**:
To find the resultant displacement \( R \) of \( x_1 \) and \( x_2 \), we use the formula for the resultant of two vectors:
\[ R = \sqrt{A^2 + A^2 + 2A \cdot A \cdot \cos\left(\frac{2\pi}{3}\right)} \]
Since \( \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \):
\[ R = \sqrt{A^2 + A^2 - A^2} = \sqrt{A^2} = A \]

4. **Determine the Angle of the Resultant**:
The angle \( \theta \) of the resultant can be calculated using the sine rule or by vector addition:
\[ \tan(\theta) = \frac{A \sin(0) + A \sin\left(\frac{2\pi}{3}\right)}{A \cos(0) + A \cos\left(\frac{2\pi}{3}\right)} \]
Since \( \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \) and \( \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \):
\[ \tan(\theta) = \frac{0 + A \cdot \frac{\sqrt{3}}{2}}{A - A \cdot \frac{1}{2}} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \]
Thus, \( \theta = \frac{\pi}{3} \) (or \( 60^\circ \)).

5. **Determine the Third Displacement \( x_3 \)**:
To bring the mass to rest, the third displacement \( x_3 \) must be equal in magnitude and opposite in direction to the resultant of \( x_1 \) and \( x_2 \):
\[ x_3(t) = B \sin\left(\omega t + \phi\right) \]
The angle \( \phi \) must be:
\[ \phi = \theta + \pi = \frac{\pi}{3} + \pi = \frac{4\pi}{3} \]
The amplitude \( B \) must be equal to the magnitude of the resultant:
\[ B = A \]

### Final Values:
Thus, the values of \( B \) and \( \phi \) are:
\[ B = A, \quad \phi = \frac{4\pi}{3} \]