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Two simple harmonic motions are given by x_(1) = a sin omegat + a cos omega t and x_(2) = asin omegat + a/sqrt(3) 01. The ratio of the amplitudes of first and second motion and the phase difference between them are respectively |
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Answer» `sqrt(3/2)` and `pi/12` `x_(1) = a sin omega t + a cos omegat` `=sqrt(2)a [1/sqrt(2) sin omegat + 1/sqrt(2) cos omegat]` `=sqrt(2)a sin (omegat + pi/4)` `THEREFORE` Amplitude of this simple harmonic motion is, `a_(1) = sqrt(2)a` and phase of the simple harmonic motion is, `phi_(1) =pi/4` For second simple harmonic motion, `x_(2) =a sin omegat + a/sqrt(3) cos omegat` `=(2a)/sqrt(3)[sqrt(3)/2 sin omegat + 1/2 cos omegat]` `(2a)/sqrt(3)sin(omegat + pi/6)` `therefore` Amplitude of this simple harmonic motion is, `a_(2) =(2a)/sqrt(3)` and phase of this simple harmonic motion is, `phi_(2) = pi/6` The ratio of their amplitudes is `a_(1)/a_(2) =(sqrt(2)a)/((2a)/sqrt(3)) = sqrt(3/2)` Phase difference between them is, `Deltaphi = phi_(1) - phi_(2) = pi/4 - pi/6 =(6pi - 4PI)/24 = (2pi)/24 = pi/12` |
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