NO

how do u get the answer

Excuse me aditi. But i am correct. The reason:

Well, the two particles equations can be written as 20cos(pT) and 20cos(pt+c) where c is the difference in phase constant. I would like to take p as 1 since it will NOT affect the answer. So, here we go!

We first have to add the equations and equate it to 20. Reason: You can get a distance of 20 only if they are on the opposite sides of the mean position. You cannot get a distance of 20 between them when they are on the same side of the mean position because that would mean when cos(T) is zero then cos(T+c) is 1 or vice-versa. Therefore that is ruled out. therefore we have to add the equations and solve for c. We will get 2pi/3. Therefore, i am correct.

Cheers,

Surya

but in special case pi/2 is also aright ans

as in the ques he doesnt mention the initial position  so one particle can be at equilibrium position and other can be in extreme     i think

A=20cm

y1=A sin(wt)  and y2=A sin(wt +c) where c is the phase difference b/w the two waves

Now as the max distance b/w the two is 20 cm then the diff in their positions will be 20 cm ( sorry surya bt i dun think u shud add d equations and equate to 20 w/e be their positions w.r.t mean position)

so y2-y1 = A ( sin(wt) - sin(wt +c) )   i.e. sin(wt +c) - sin(wt) = 1

as the max distance is equal to A . so this will be eqn 1

Now as they are at maximum distance their velocity should be same so second eqn will be v1=v2

i.e. Awcos wt = Aw cos(wt+c)

i.e. cos wt = cos (wt+c)

or, wt = 2n(pi) + or -(wt+c)

we get c = 0 or c=(-2wt)

but c=0 doesnt satisfy eqn 1

so putting c = -2wt in eqn 1 we get c = pi/3  which is d Answer

--Aditi

why shud vel be same at max dist?

yes, really sorry.....wrote the reason wrong

It wudn't be correct to say that they wud have d same velocity at the max distance as it cud be anything, nothing can be said bout the velocity i guess !

But the velocity eqn i.e. the second one, we get dat by differentiating the eqn 1 i.e. the displacement eqn !!! not by equating the velocities, ohh no, even though that also gives the same eqn but still the reason will be wrong !

how can u say so ? the velocity of the particles at the moment of maximum displacement can be anything as they dun both start at the same time !

that need not be true einstein..the two bodies can be anywhere

excuse me aditi akka, but i have a valid reason that we should add the two. Suppose we take the case of your method, then when you substract y2 and y1 you imply that the two particles are on the same side of their mean position. And you imply that the distance between them is 20. But the maximum extent to which the particles move is 20cm from the mean position so that means that one particle is at the mean position and one is at the extreme. But if you look at the four graphs of cosx, cosx+pi/4 , cosx+pi/3, cosx+2pi/3,etc you will never get one equations is zero and the other is plus or minus one except for cosx+2pi/3. That's why your method is wrong. I saw many of my friends commit the mistake you have and told my teacher to rectify in everyone's textbooks. that is why i'm confident.

Don't take this harshly ok aditi akka?

cheers,

surya