Phase Difference Homework: Find Magnitude of Phase Difference at Detector

[KingBigness]

Homework Statement

Sources A and B emit long-range radio waves of wavelength 550 m, with the phase of the emission from A ahead of that from source B by 90°. The distance rA from A to a detector is greater than the corresponding distance rB from B by 140 m. What is the magnitude of the phase difference at the detector?

Homework Equations

The Attempt at a Solution

Initially, source A leads source B by 90°, which is equivalent to \frac{1}{4} wavelength.
However, source A also lags behind source B since rA is longer than rB by 140 m, which is \frac{140}{550}=0.2545 wavelength. So the net phase difference between A and B at the
detector is -0.732 Rad.

This is wrong though, can someone help me =D

 

[vela]

How'd you come up with -0.732 rad? It looks like you completely ignored one of the terms.

 

[KingBigness]

To be honest I'm not sure.
I have tried everything I can think off but all my answers are wrong.

 

[KingBigness]

Thank you for your time but I asked a friend and he showed me. It was just a stupid mistake that I missed thank yiu

 

[asteriatic]

Hello,

To find the magnitude of the phase difference at the detector, we can use the formula:

Phase difference = (2π/λ) * (rA - rB)

Where λ is the wavelength (550 m in this case), and rA and rB are the distances from sources A and B respectively to the detector.

Substituting the given values, we get:

Phase difference = (2π/550) * (rA - rB)

= (2π/550) * (rA - (rA - 140))

= (2π/550) * 140

= 0.4036 radians

Therefore, the magnitude of the phase difference at the detector is 0.4036 radians.