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.
To solve for AC, we can manipulate the equation:
AC = BC + 2.54/2
We can substitute the value of BC from the given equation BC = 5.8 - AC:
AC = (5.8 - AC) + 2.54/2
Now, simplify the equation:
2AC = 5.8 - AC + 2.54/2
Multiply all terms by 2 to remove the fraction:
4AC = 11.6 - 2AC + 2.54
Rearrange the equation:
6AC = 11.6 + 2.54
Combine the two constants:
6AC = 14.14
Divide both sides by 6 to solve for AC:
AC = 14.14 / 6
AC ≈ 2.3567
So, the value of AC is approximately 2.3567.