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Phase difference between the particles

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Two particles P and Q describe SHM of same amplitude A, frequency f along the same straight line. The maximum distance between the two particles is $A\sqrt2$. The initial phase difference between the particles is ?

What I did is that thought that both the particles if were at extreme ends initially then maximum distance would have been 2A ; and phase difference should be $\pi$

Then if one was at extreme and other at mean initially then difference would have been A; and phase difference would be $\pi/2$

But given is between some where between 2A and A i.e. $A\sqrt2$ ,

How should I take out phase difference for that ?

asked Apr 1, 2017 in Physics Problems by physicsapproval (2,290 points)

1 Answer

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Best answer

Your reasoning is correct, but you need to go a little further to deduce the answer.

The question does not say but we must assume that the mean position O is the same for P and Q. The phase difference between them remains the same, but the separation changes.

Suppose that, at some instant, P and Q are moving in the same direction. When the separation is maximum they are moving with the same speed. If the speed of one was greater than the other, they would get closer together of further apart. For each, speed decreases with distance from O in the same manner, because they have the same amplitude. Whichever is closer to O moves faster. They will have the same speed when they are the same distance from O. So at maximum separation they will be positioned symmetrically about O.

If the maximum separation is $A\sqrt2$ then each is $\frac{A\sqrt2}{2}=\frac{A}{\sqrt2}$ from O. The phase of each is then $\phi$ where
$A\sin\phi=\frac{A}{\sqrt2}$
$\sin\phi=\frac{1}{\sqrt2}=\sin45^{\circ}$
$\phi=45^{\circ}$.

So the phase difference at maximum separation (and at all separations) is $2\phi=90^{\circ}$.


Alternative solution :

$x_1=A\sin(\omega t+\phi)$
$x_2=A\sin(\omega t)$
$x_1-x_2=2A\cos(\frac{\omega t+\phi+\omega t}{2})\sin(\frac{\omega t+\phi - \omega t}{2})=2A\cos(\omega t+\frac12 \phi)\sin(\frac12 \phi)$.

$\sin(\frac12 \phi)$ is a constant, so the maximum possible value of $x_1-x_2$ occurs when $\cos(\omega t+\frac12 \phi)=1$. Then
$2A\sin(\frac12 \phi)= A\sqrt2$
$\sin(\frac12 \phi)=\frac{\sqrt2}{2}=\frac{1}{\sqrt2}=\sin 45^{\circ}$
$\phi=90^{\circ}$.

answered Apr 1, 2017 by sammy gerbil (26,096 points)
selected Apr 1, 2017 by physicsapproval
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