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"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
3 weeks, 6 days ago

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