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

Explanation

To solve the problem, we need to analyze the given sinusoidal displacements and find the values of \( B \) and \( \phi \) such that the total displacement brings the mass to rest. Here’s a step-by-step solution:

### Step 1: Understanding the Given Displacements
We have two sinusoidal displacements:
1. \( x_1(t) = A \sin(\omega t) \)
2. \( x_2(t) = A \sin\left(\omega t + \frac{2\pi}{3}\right) \)

### Step 2: Adding the Two Displacements
We need to find the resultant of \( x_1 \) and \( x_2 \). Since both displacements have the same amplitude \( A \) and a phase difference of \( \frac{2\pi}{3} \) (or 120 degrees), we can use the formula for the resultant of two vectors with the same magnitude:

\[ R = 2A \cos\left(\frac{\theta}{2}\right) \]

where \( \theta = \frac{2\pi}{3} \).

### Step 3: Calculating the Resultant Amplitude
Calculating \( R \):
\[ R = 2A \cos\left(\frac{\frac{2\pi}{3}}{2}\right) = 2A \cos\left(\frac{\pi}{3}\right) \]
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]
Thus,
\[ R = 2A \cdot \frac{1}{2} = A \]

### Step 4: Finding the Phase of the Resultant
The resultant phase angle \( \phi_R \) can be found using the angle bisector theorem:
\[ \phi_R = \frac{\theta}{2} = \frac{120^\circ}{2} = 60^\circ \text{ or } \frac{\pi}{3} \text{ radians} \]
So, the resultant displacement can be expressed as:
\[ x_R(t) = A \sin\left(\omega t + \frac{\pi}{3}\right) \]

### Step 5: Adding the Third Displacement
Now, we introduce the third displacement:
\[ x_3(t) = B \sin(\omega t + \phi) \]
We need the total displacement to be zero:
\[ x_1 + x_2 + x_3 = 0 \]
This implies:
\[ A \sin\left(\omega t + \frac{\pi}{3}\right) + B \sin(\omega t + \phi) = 0 \]

### Step 6: Setting Up the Equation
For the total to be zero, \( x_3(t) \) must be equal to the negative of the resultant:
\[ x_3(t) = -A \sin\left(\omega t + \frac{\pi}{3}\right) \]

### Step 7: Comparing Amplitudes and Phases
From this, we can see that:
1. The amplitude \( B \) must equal \( A \):
\[ B = A \]
2. The phase \( \phi \) must be:
\[ \phi = \frac{\pi}{3} + \pi = \frac{4\pi}{3} \]

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