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

Step 1: Represent the three SHMs as phasors of equal length \(A\) at phases \(0^\circ\), \(60^\circ\), and \(120^\circ\).

Step 2: Add the horizontal (x) components:\[A_x = A(\cos 0^\circ + \cos 60^\circ + \cos 120^\circ) = A\left(1 + \tfrac{1}{2} - \tfrac{1}{2}\right) = A.\]
Step 3: Add the vertical (y) components:\[A_y = A(\sin 0^\circ + \sin 60^\circ + \sin 120^\circ) = A\left(0 + \tfrac{\sqrt{3}}{2} + \tfrac{\sqrt{3}}{2}\right) = \sqrt{3}\,A.\]
Step 4: The resultant amplitude is\[A_R = \sqrt{A_x^2 + A_y^2} = \sqrt{A^2 + 3A^2} = \sqrt{4A^2} = 2A.\]This is option (B).\[\boxed{A_R = 2A}\]