Question

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. The amplitude of the resultant motion will be:

• A. \(A\)
• B. \(2A\)
• C. \(\sqrt{2}\,A\)
• D. \(3A\)

Hint:

Add the three equal phasors at \(0^\circ, 60^\circ, 120^\circ\). Sum the x and y parts, then take \(\sqrt{A_x^2 + A_y^2}\).

Answer 1

The Correct Option is B

Draw the three equal phasors head to tail, each of length $A$, turning by $60^\circ$ each time: directions $0^\circ$, $60^\circ$, $120^\circ$.

A neat symmetry helps here. The first ($0^\circ$) and third ($120^\circ$) phasors are placed symmetrically about the middle one ($60^\circ$). Their vector sum therefore lies along the $60^\circ$ direction. Each of them makes $60^\circ$ with the middle phasor, so each contributes $A\cos 60^\circ = \tfrac{A}{2}$ along that middle direction, together giving $A$.

Adding the middle phasor itself, which already points along $60^\circ$ with length $A$, the total along the $60^\circ$ line is\[A + A = 2A.\]The perpendicular contributions of the outer two cancel by symmetry, so the resultant amplitude is simply\[A_R = 2A.\]\[\boxed{A_R = 2A}\]