Step-by-Step Solution

1. Understanding the Condition for Rest

The problem states that adding a third sinusoidal displacement, $x_3(t)$, brings the mass to a complete rest. This means the net displacement of the mass is zero for all time $t$. Mathematically, this condition is expressed as: $x_{net}(t) = x_1(t) + x_2(t) + x_3(t) = 0$ This must hold true for any value of $t$. This implies that the third displacement must be the negative of the sum of the first two displacements: $x_3(t) = - (x_1(t) + x_2(t))$

2. Using the Phasor Method

Simple Harmonic Motions (SHMs) of the same frequency can be represented by rotating vectors called phasors. The condition that the net displacement is zero is equivalent to the condition that the vector sum of the phasors is zero.

• The first displacement $x_1(t) = A ext{sin}(\omega t)$ can be represented by a phasor $\vec{P_1}$ of magnitude $A$ at an angle of $0$ radians with respect to the positive x-axis.
• The second displacement $x_2(t) = A ext{sin}(\omega t + \frac{2\pi}{3})$ is represented by a phasor $\vec{P_2}$ of magnitude $A$ at an angle of $\frac{2\pi}{3}$ radians.
• The third displacement $x_3(t) = B ext{sin}(\omega t + \phi)$ is represented by a phasor $\vec{P_3}$ of magnitude $B$ at an angle of $\phi$.

The condition for the mass to be at rest is: $\vec{P_1} + \vec{P_2} + \vec{P_3} = \vec{0}$ This implies that: $\vec{P_3} = - (\vec{P_1} + \vec{P_2})$

3. Finding the Resultant of the First Two Displacements

Let's find the resultant phasor $\vec{R}_{12} = \vec{P_1} + \vec{P_2}$. We can do this by adding their components.

• x-component of $\vec{R}_{12}$: $R_{12,x} = A ext{cos}(0) + A ext{cos}\left(\frac{2\pi}{3}\right) = A + A\left(-\frac{1}{2}\right) = \frac{A}{2}$

• y-component of $\vec{R}_{12}$: $R_{12,y} = A ext{sin}(0) + A ext{sin}\left(\frac{2\pi}{3}\right) = 0 + A\left(\frac{\sqrt{3}}{2}\right) = \frac{A\sqrt{3}}{2}$

Now, we find the magnitude and phase of the resultant phasor $\vec{R}_{12}$.

• Magnitude of $\vec{R}_{12}$: $|\vec{R}_{12}| = \sqrt{R_{12,x}^2 + R_{12,y}^2} = \sqrt{\left(\frac{A}{2}\right)^2 + \left(\frac{A\sqrt{3}}{2}\right)^2} = \sqrt{\frac{A^2}{4} + \frac{3A^2}{4}} = \sqrt{\frac{4A^2}{4}} = \sqrt{A^2} = A$

• Phase angle ($\ heta$) of $\vec{R}_{12}$: $\tan(\theta) = \frac{R_{12,y}}{R_{12,x}} = \frac{A\sqrt{3}/2}{A/2} = \sqrt{3}$ Since both x and y components are positive, the angle is in the first quadrant. So, $\theta = \frac{\pi}{3}$.

Thus, the resultant of the first two displacements is $x_{12}(t) = A\sin(\omega t + \frac{\pi}{3})$.

4. Determining B and $\phi$ for the Third Displacement

From step 2, we have $\vec{P_3} = -\vec{R}_{12}$. This means the phasor for the third displacement must have the same magnitude as $\vec{R}_{12}$ but be in the opposite direction (a phase difference of $\pi$ radians).

• The magnitude $B$ of $\vec{P_3}$ is equal to the magnitude of $\vec{R}_{12}$: $B = |\vec{R}_{12}| = A$

• The phase angle $\phi$ of $\vec{P_3}$ is the phase of $\vec{R}_{12}$ plus $\pi$: $\phi = \theta + \pi = \frac{\pi}{3} + \pi = \frac{4\pi}{3}$

Therefore, the values for the third displacement are $B=A$ and $\phi = \frac{4\pi}{3}$.

5. Comparing with Options

The calculated values are $B=A$ and $\phi = \frac{4\pi}{3}$. This matches option B.

Final Answer: The correct option is B: $A,{{4\pi } \over 3}$.