Three harmonic oscillations of the same direction are added together. Their equations are given by:

Question

Three harmonic oscillations of the same direction are added together. Their equations are given by: $\xi_1 = a \cos \omega t$, $\xi_2 = 2a \sin \omega t$, and $\xi_3 = 1.5a \cos (\omega t + \pi/3)$. The amplitude of the resultant oscillation is closest to:

Professor PC Thomas Classes · Answer

Let the resultant be $A \cos(\omega t + \phi)$. We can convert all terms to cosines to use phasors: $\xi_1 = a \cos \omega t$, $\xi_2 = 2a \cos(\omega t - \pi/2)$, $\xi_3 = 1.5a \cos(\omega t + \pi/3)$. The x and y components of the phasors (taking $\xi_1$ along the x-axis) are: $A_x = a + 2a \cos(-\pi/2) + 1.5a \cos(\pi/3) = a + 0 + 1.5a(0.5) = 1.75a$. $A_y = 0 + 2a \sin(-\pi/2) + 1.5a \sin(\pi/3) = -2a + 1.5a(\sqrt{3}/2) = (-2 + 0.75\sqrt{3})a \approx -0.701a$. The resultant amplitude $A = \sqrt{A_x^2 + A_y^2} = a \sqrt{(1.75)^2 + (-0.701)^2} = a \sqrt{3.0625 + 0.4914} = a \sqrt{3.5539} \approx 1.88a$.