Wave and Particle

Question

The velocity of point P when its dispalcement is !!4cm;!!

(a)!!(3sqrt3 pi)/125 hati m//s!!

(b) !!-(3sqrt3 pi)/125 hati m//s!!

(c) !!(3sqrt3 pi)/125 hatj m//s!!

(d) !!-(3sqrt3 pi)/125 hatj m//s!!

Solution

The point P moves in positive !!y!!-direction as the wave moves in positive !!x!!-direction. So it's velocity is positive.

Let the position of point P with be represented by,

!!y = A sin(omega t+-phi).....(1)!!

Differentiate with respect to !!t!!, we get

!!dy/(dt)= v = Aomega cos(omega t +-phi).....(2)!!

At time !!t!!, we know that !!y=4cm!!

Putting the value of !!y!! in equation !!(1)!!, we get

!! 4 = 8 sin(omega t+-phi)=> sin(omega t+-phi)= 1/2!!

!!=> omega t+-phi= pi/6!!

We know that,

!!omega = (2piv(wave))/lambda!!

!!=> omega = (2pixx0.15)/(0.5)= (3pi)/5!!

Putting the values in equation !!(2)!!,

!!v = (8/100)xx((3pi)/5)xx cos(pi/6)!!

!!=> v = (3sqrt3 pi)/125 hatj m//s!!

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