What is the Phase Difference Between Waves at Point R?

AI Thread Summary
The discussion focuses on determining the phase difference between two coherent wave sources, P and Q, at point R. It is established that if P and Q are equidistant from R, the phase difference is zero; however, the user notes that PR and QR are not equal, complicating the calculation. The user attempts to derive the equations for the waves at P and Q and combines them using the principle of superposition. The resulting expression indicates that the phase difference can be derived from the distance difference between the two waves. The conversation emphasizes the need to accurately assess the distances to find the correct phase difference at point R.
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Homework Statement



The diagram attached shows two coherent sources P and Q which produce waves of the same phase with wavelength , labda . If the two waves meet at the point R , the phase difference between waves at R is ??

Homework Equations



y=a sin (omega t + kx)

The Attempt at a Solution



i have problems obtaining the equation for wave p and q respectively , i know i have to find the superposition of the wave ?
 

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If P and Q are equidistant from R, then the phase difference between the waves at R is equal to zero.
 


rl.bhat said:
If P and Q are equidistant from R, then the phase difference between the waves at R is equal to zero.

but PR is not equal to QR , which part of the question says they are equal ?
 


If PR = x1 and QR = x2, then
y1 = a*sin(ωt + kx1)
Y2 = a*sin(ωt + kx2)
y1 + y2 = a*[sin(ωt+kx1) + sin(ωt+kx2)]
= a*2*sin[ωt + k(x1+x2)/2]*cos[(x1-x2)/2]
= 2*a*cos[(x1-x2)/2]*sin[ωt + k(x1+x2)/2]
Now find the phase difference.
 


rl.bhat said:
If PR = x1 and QR = x2, then
y1 = a*sin(ωt + kx1)
Y2 = a*sin(ωt + kx2)
y1 + y2 = a*[sin(ωt+kx1) + sin(ωt+kx2)]
= a*2*sin[ωt + k(x1+x2)/2]*cos[(x1-x2)/2]
= 2*a*cos[(x1-x2)/2]*sin[ωt + k(x1+x2)/2]
Now find the phase difference.

thanks !
 
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