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?

Question

anonymous · Answer

$$X=Asin \theta $$

anonymous · Answer

$$\frac{ A }{ 2 }=Asin \theta $$

abhisar · Answer

Yes, 

$\sf \phi_1$ will be $\sf \large\frac{\pi}{6}$

abhisar · Answer

Next?

anonymous · Answer

$$\theta=30^o ,150^o $$

anonymous · Answer

difference will be $$150^o -30^o =120^0 \rightarrow \frac{ 2\pi }{ 3 }$$

abhisar · Answer

Yes, can you explain why $\sf \theta_2$ will be 150?

anonymous · Answer

because the two particles are moving in opposite directions so their respective points on the wheel described above will be at exactly opposite points on the wheel.

abhisar · Answer

Show me how you got 150 using the formula you mentioned above.

abhisar · Answer

X=ASin$\sf \theta$

anonymous · Answer

$$\phi=\pi -\frac{ \pi }{ 6 }=\frac{ 5\pi }{ 6 }$$

anonymous · Answer

this is easier way

anonymous · Answer

because opposite just subtract by 180

anonymous · Answer

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anonymous · Answer

two rotating

anonymous · Answer

@ganeshie8  do u mind explaining , i think i a not explaining it right?

abhisar · Answer

@ganeshie8 Hi!

I just need to understand one thing...

abhisar · Answer

I know how to solve it, 

Phase for second particle will be $\sf \pi -\frac{\pi}{6}$

Why we subtract it from $\sf \pi$?

ganeshie8 · Answer

motion of particle 1 is represented as phasor in below diagram 
Lets assume that the particle is oscillating back and forth along y axis and the component of phasor along y axis gives us the displacement of the particle :

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