1.

Two particles execute SHM of same amplitude and frequency on parallel lines. They cross each another when moving in opposite directions each time their displacement is half their amplitude. What is the phase difference between them?

Answer»

Solution :If we assume that the particles are initially at the MEAN position, their equationfor displacement.
`x=A sin omega t`
But `x=A/2"":.A/2=A sin omega`(or `sin omegat=1/2`
PHASE `=omegat=30^(@),150^(@)`
`( :. sin (180^(@)-theta)=sin theta, sin (180^(@)=30^(@))=sin 30^(@))`
ONe of the particles has phase of `30^(@)` and the other has phase of `150^(@)`
Phase different between them `=120^(@)=(2pi)/3` radian


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