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Two particles execute simple harmonic motion of same amplitude and frequency along the same straight line. They pass on another, when going in opposite directions, each time their displacement is half of their amplitude. What is the phase difference between them?A. `120^(@)`B. `180^(@)`C. `60^(@)`D. `30^(@)` |
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Answer» Correct Answer - A Let `theta` be the phase difference betweent two simple harmonic motions, r and `omega` be the amplitude and angular frequency of motion of the particles. The two SHMs are represented by the equations. `y_(1)=rsin omegat and y_(2)=rsin (omegat+theta)`. Let the two particles cross one another, moving in opposite directions at time `t=t_(1)` (say) when displacement of each particle is `r//2`, then `y_(1)=y_( 2)=r//2` at time ` t=t_(1)`. `:. (r)/(2)=rsin omegat_(1) or sinomegat_(1)=(1)/(2):. cos omega t_(1)=(sqrt(3))/(2)` Also `(r)/(2)=rsin(omegat_(1)+theta) or (1)/(2)=sin (omegat_(1)+theta)` `(1)/(2)=sin omegat_(1)costheta+cosomegat_(1)sintheta` or `(1)/(2)=(1)/(2)costheta+(sqrt(3))/(2)sintheta` or `1-costheta+sqrt(3) sin theta or 1-cos theta =sqrt(3) sin theta` squaring both sides, we get `1+cos^(2)theta-2cos theta=3sin^(2)theta=3(1-cos^(2)theta)` `4cos^(2)theta-2cos theta-2=0` or `(4cos theta +2)(cos theta-1)=0` when `4cos theta -1=0, cos theta =1, theta=0^(@)`, this is true when both particles start together. `:. `Phase difference, `theta=120^(@)` |
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