Three simple harmonic motions of equal amplit | vedclass

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(B) Let the three SHMs be represented by phasors of magnitude $A$ at angles $0^{\circ}$,$60^{\circ}$,and $120^{\circ}$ with respect to the first motio

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$2\,A$

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(B) Let the three SHMs be represented by phasors of magnitude $A$ at angles $0^{\circ}$,$60^{\circ}$,and $120^{\circ}$ with respect to the first motion.The resultant amplitude $A_{	ext{net}}$ is given by the vector sum of these three phasors:$A_{	ext{net}} = \sqrt{(A + A\cos 60^{\circ} + A\cos 120^{\circ})^2 + (0 + A\sin 60^{\circ} + A\sin 120^{\circ})^2}$$A_{	ext{net}} = \sqrt{(A + A/2 - A/2)^2 + (0 + A\sqrt{3}/2 + A\sqrt{3}/2)^2}$$A_{	ext{net}} = \sqrt{A^2 + (A\sqrt{3})^2} = \sqrt{A^2 + 3A^2} = \sqrt{4A^2} = 2A$Alternatively,using the phasor diagram,the resultant of the first and third phasors (which are at $120^{\circ}$ to each other) is a vector of magnitude $A$ directed along the same line as the second phasor. Adding this to the second phasor gives $A + A = 2A$.

Three simple harmonic motions of equal amplitudes $A$ and equal time periods in the same direction combine. The phase of the second motion is $60^{\circ}$ ahead of the first and the phase of the third motion is $60^{\circ}$ ahead of the second. Find the amplitude of the resultant motion.

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