Answer to a Question Recently Posed

Someone asked this question recently:

"Two particles are simultaneously performing SHM along the same path with same time period and equal amplitude A. If the maximum separation b/w particles if (root3)A, then find their phase difference and positions when they cross each other."

This question involves some interesting mathematics. Consider the complex representation of a sinusoid ("Re" denotes "real part" and "j" is the imaginary unit):

\[A \, cos(\omega t) = Re[A e^{j \omega t}]\]

Notice that the coefficient on the exponential term is the magnitude of the sinusoid. That will be important to remember. Write general expressions for the two sinusoids under consideration:

\[f_1 = A \, cos(\omega t) \\ f_2 = A \, cos(\omega t + \theta) \]

Take the difference:

\[f_1 - f_2 = A \, cos(\omega t) - A \, cos(\omega t + \theta) \\ = Re[A e^{j \omega t}] - Re[A e^{j (\omega t + \theta)}] \\ = Re[A e^{j \omega t} - A e^{j (\omega t + \theta)} ] \\ = Re[e^{j \omega t} (A - A e^{j \theta}) ]\]

We can therefore infer that the magnitude (peak value) of the resultant sine wave is the magnitude of the complex quantity \(A - A e^{j \theta}\), which we also know to be \(\sqrt{3} A\). The peak value of the resultant sine wave will occur when the complex exponential has a phase angle that is the negation of the phase angle of \(A - A e^{j \theta}\), making the argument of the Re() operation simply equal to the length of \(A - A e^{j \theta}\). The following therefore must be true:

\[|1 - e^{j \theta}|^2 = 3 \\ (1- cos\theta)^2 + (sin\theta)^2 = 3 \\ -2 cos\theta = 1 \\ \implies cos\theta = -\frac{1}{2}\]

The two sinusoids are therefore separated by plus or minus 120 degrees. The other timing information should be fairly easy to figure out from here.

Note by Steven Chase
10 months, 4 weeks ago

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