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):

Acos(ωt)=Re[Aejωt]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:

f1=Acos(ωt)f2=Acos(ωt+θ)f_1 = A \, cos(\omega t) \\ f_2 = A \, cos(\omega t + \theta)

Take the difference:

f1f2=Acos(ωt)Acos(ωt+θ)=Re[Aejωt]Re[Aej(ωt+θ)]=Re[AejωtAej(ωt+θ)]=Re[ejωt(AAejθ)]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 AAejθA - A e^{j \theta}, which we also know to be 3A\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 AAejθA - A e^{j \theta}, making the argument of the Re() operation simply equal to the length of AAejθA - A e^{j \theta}. The following therefore must be true:

1ejθ2=3(1cosθ)2+(sinθ)2=32cosθ=1    cosθ=12|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
1 year, 8 months ago

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