A point mass is subjected to two simultaneous | vedclass

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(B) The resultant of the first two displacements is given by $x_1 + x_2 = A \sin \omega t + A \sin \left(\omega t + \frac{2 \pi}{3}\right)$.Using the

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$A, \frac{4 \pi}{3}$

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(B) The resultant of the first two displacements is given by $x_1 + x_2 = A \sin \omega t + A \sin \left(\omega t + \frac{2 \pi}{3}ight)$.Using the trigonometric identity $\sin C + \sin D = 2 \sin \left(\frac{C+D}{2}ight) \cos \left(\frac{C-D}{2}ight)$,we get:$x_1 + x_2 = 2A \sin \left(\omega t + \frac{\pi}{3}ight) \cos \left(-\frac{\pi}{3}ight) = 2A \sin \left(\omega t + \frac{\pi}{3}ight) \cdot \frac{1}{2} = A \sin \left(\omega t + \frac{\pi}{3}ight)$.For the mass to be at complete rest,the sum of all displacements must be zero: $x_1 + x_2 + x_3 = 0$.Therefore,$x_3 = -(x_1 + x_2) = -A \sin \left(\omega t + \frac{\pi}{3}ight)$.Using the identity $-\sin 	heta = \sin (	heta + \pi)$,we get:$x_3 = A \sin \left(\omega t + \frac{\pi}{3} + \piight) = A \sin \left(\omega t + \frac{4 \pi}{3}ight)$.Comparing this with $x_3(t) = B \sin (\omega t + \phi)$,we find $B = A$ and $\phi = \frac{4 \pi}{3}$.

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