f=140 Hz
v=343
wavelength=(v/f)=(343/140)=2.45
for destructive interference=(wavelength/2)=(2.45/2)=1.225
Solve for x which is the distance between speaker B and the observer C.
By pythagorean, CB^2+BC^2=AC^2
x^2+4.8^2=AC^2
AC^=(x+1.225)^2
So, we have
x^2+4.8^2=(x+1.225)^2 factor
x^2+4.8^2=x^2+1.225x+1.225x+1.225^2
x^2+4.8^2=x^2+2.45x+1.5
23.04=2.45x+1.5
21.54=2.45x
x=8.792
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
For Further Reading
Physics, please help - bobpursley, Friday, May 4, 2007 at 11:54am
AC=BC + 2.54/2 where BC= 5.8-AC
AC=5.8-AC + 2.54/2
2AC= ... solve for AC.
I just want to understand this. How come it's bc= 5.8 - AC
Using AC= sqrt [ (AB)^2 + (BC)^2]
than AC = 5.8 + BC
BC = AC - 5.8
That bc=5.8-AC is when C is on the line AB, that is, inbetween two speakers. The problem was unclear on that as I recall. Can you repost the original question?
suppose that the separation between speakers A and B is 4.80 m and the speakers are vibrating in phase. They are playing identical 140 Hz tones, and the speed of sound is 343 m/s. What is the largest possible distance between speaker B and the observer at C, such that he observes destructive interference
A B
speaker......................speaker-
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observer -
Ok. The largest distance is infinity. Destructive interference will occur then the path length from B to O is and odd 1/2 wavelengths longer than A to O.
That occurs for any multiples of the odd number of half wavelengths.
could u please show ur work?
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