Superposition of Sinusoidal Displacements for Mass at Rest

To bring the point mass to a complete rest, the resultant displacement of all three superimposed sinusoidal displacements must be identically zero at all times $t$:

$x_1(t) + x_2(t) + x_3(t) = 0$

Given the individual displacements:
$x_1(t) = A \sin \omega t$
$x_2(t) = A \sin\left(\omega t + \frac{2\pi}{3}\right)$
$x_3(t) = B \sin(\omega t + \phi)$

First, let us combine the displacements $x_1(t)$ and $x_2(t)$ using the trigonometric sum-to-product identity $\sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$:

$x_1(t) + x_2(t) = A \left[ \sin \omega t + \sin\left(\omega t + \frac{2\pi}{3}\right) \right]$

$x_1(t) + x_2(t) = 2A \sin\left(\omega t + \frac{\pi}{3}\right) \cos\left(-\frac{\pi}{3}\right)$

Since $\cos\left(-\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$, we get:

$x_1(t) + x_2(t) = 2A \cdot \frac{1}{2} \cdot \sin\left(\omega t + \frac{\pi}{3}\right) = A \sin\left(\omega t + \frac{\pi}{3}\right)$

Now, substituting this back into the condition for complete rest:

$A \sin\left(\omega t + \frac{\pi}{3}\right) + B \sin(\omega t + \phi) = 0$

$B \sin(\omega t + \phi) = -A \sin\left(\omega t + \frac{\pi}{3}\right)$

Using the identity $-\sin(\theta) = \sin(\theta + \pi)$:

$B \sin(\omega t + \phi) = A \sin\left(\omega t + \frac{\pi}{3} + \pi\right)$

$B \sin(\omega t + \phi) = A \sin\left(\omega t + \frac{4\pi}{3}\right)$

Equating the amplitude $B$ and phase $\phi$ on both sides:

$B = A$
$\phi = \frac{4\pi}{3}$

Hence, the correct option is B.