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  • Solve it, Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean posit

Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is

  • Option 1)

    0

  • Option 2)

    1

  • Option 3)

    \pi

  • Option 4)

    \pi/6

 

Answers (1)

best_answer

As discussed

Relation between phase velocity and wave speed -

V_{P}= -V: frac{dy}{dx}

- wherein

V_{P}= particle velocity

V = wave velocity

frac{dy}{dx}= slope of curve

 

 y_{1}=a Sin (wt+ICx+0.57)

\phi =wt+ICx+0.57

Y_{2} =aCos (wt+Kx) =a Cos (wt +Kx)=a Cos (wt+Kx+\frac{\pi}{2})\phi _{2} = wt+Kx+\frac{\pi}{2}

\Delta\phi=\phi_{2} - \phi_{1} = (wt+kx-\frac{\pi}{2}) - (wt+kx+0.57)

=\frac{\pi}{2} -0.57 = (1.57-0.57) radius

=1 radius 


Option 1)

0

Option 2)

1

Option 3)

\pi

Option 4)

\pi/6

Posted by

prateek

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