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Distruive interference? Unless you put C in the line between A and B, this the asnwer is at infinity.

Now on the line between A and B, the distance from A to C has to be an odd multiple of halfwavelength greater than C to B. Now, the distance between the speakers is two wavelength. So if AC-BC is n lambda/2, then n must be one, so
AC=BC + 2.54/2 where BC= 5.8-AC

solve for AC.

Physics - Mary, Tuesday, May 1, 2007 at 9:37pm
Please tell me where I am going wrong.

AC = BC + 2.54/2

AC = 7.366

AC=BC + 2.54/2 where BC= 5.8-AC
AC=5.8-AC + 2.54/2
2AC= ... solve for AC.

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