Waves and Sound Waves and Sound Physics Help Forum 
May 7th 2011, 08:52 AM

#1  Junior Member
Join Date: Apr 2008
Posts: 6
 Resonance Tube
Question:
In a resonance tube, the first vibrating length is 0.6m. The second resonating length is 0.62m. Find the inner diameter of the tube.
Options:
a) 6.66cm
b) 4.67cm
c) 3.33cm
d) 2.33cm
Attempt:
I tried finding out the end correction to find the diameter (e=0.3d, right?), but I'm getting a negative value for end correction. I'm not sure what that means. The tuning fork is inside the tube? :S
Cheers.

 
May 7th 2011, 10:22 AM

#2  Forum Admin
Join Date: Apr 2008 Location: On the dance floor, baby!
Posts: 2,084

Originally Posted by Ruler of Hell Question:
In a resonance tube, the first vibrating length is 0.6m. The second resonating length is 0.62m. Find the inner diameter of the tube.
Options:
a) 6.66cm
b) 4.67cm
c) 3.33cm
d) 2.33cm
Attempt:
I tried finding out the end correction to find the diameter (e=0.3d, right?), but I'm getting a negative value for end correction. I'm not sure what that means. The tuning fork is inside the tube? :S
Cheers. 
I'm not quite sure of the picture. I know that you can resonate on two ends if, for example, the tube is partly filled with water. In that case we'd need to know the density of the water. But what your question seems to imply is some kind of inner and outer chamber?
Or am I simply being dense and you are saying that the first mode of vibration is at 0.6 m and the second mode is at 0.62 m?
Dan
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May 7th 2011, 11:40 AM

#3  Junior Member
Join Date: Apr 2008
Posts: 6

Naw, not that complicated.
Or am I simply being dense and you are saying that the first mode of vibration is at 0.6 m and the second mode is at 0.62 m?

^^This.
Here's what it looks like:
Also, by inner diameter I mean just the diameter, as in the thickness of the tube is irrelevant. No multiple chambers or anything fancy.
Just your standard resonance tube problem.

 
May 7th 2011, 03:16 PM

#4  Forum Admin
Join Date: Apr 2008 Location: On the dance floor, baby!
Posts: 2,084

Originally Posted by Ruler of Hell Question:
In a resonance tube, the first vibrating length is 0.6m. The second resonating length is 0.62m. Find the inner diameter of the tube.
Options:
a) 6.66cm
b) 4.67cm
c) 3.33cm
d) 2.33cm
Attempt:
I tried finding out the end correction to find the diameter (e=0.3d, right?), but I'm getting a negative value for end correction. I'm not sure what that means. The tuning fork is inside the tube? :S
Cheers. 
I see now, thank you. I've never messed with the "end correction" before. That's really what ended up screwing me up. Anyway...
If I understand you then we have two equations:
The substitution method is my favorite. I solved the first equation for (lambda) and got
and plugging that into the second equation gives
That gives me a value of D = 1.967 m.
Dan
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May 8th 2011, 02:04 AM

#5  Junior Member
Join Date: Apr 2008
Posts: 6

Thanks Dan. I did something similar. Good to know I wasn't messing up.
Cheers.

 
May 21st 2011, 10:17 AM

#6  Physics Team
Join Date: Feb 2009
Posts: 1,425
 This one’s a little tricky. It hasn’t been mentioned anywhere that the resonances occur at the fundamental frequency. Let’s write lambda as Y Consider this. The end correction is usually small compared to the resonating length. So if we consider it to be resonating at the fundamental at 0.62 (since 0.62>0.6), then we should expect a wavelength lambda close to (4 x 0. 62) or 2.48. We know that for a tube closed at one end, the allowed resonances correspond to Y/4, 3Y/4 , 5Y/4 i.e. (2n+1) Y/4 ; n = 0,1,2…… The difference between successive allowed lengths thus works out to (3Y/4) – Y/4 = Y/2 = 0.62  0.6 = 0.02 or Y = 2 x 0.02 =0.04 Compare this with the earlier value. Obviously something’s wrong. What’s actually happening is this. An allowed frequency higher than the fundamental is what causes the resonance, and the wavelength corresponding to this frequency (not the fundamental wavelength) is equal to twice this difference in length, Thus if L is the effective length and L’ the measured length, we may write L = L’ + 0.3d. Since the allowed Ys are given by, (2n+1)Y/4 = L, we may write for the two cases Y = 4 L / (2n+1) = 4(L’ + 0.3d)/(2n+1). Note that the two lengths 0.62 and 0.6 correspond to the same wavelength but different numbers of this wavelength. They do not represent resonance at different frequencies. Thus we may write, 4 (0.6+0.3d )/(2n+1) = 4(0.62+0.3d)/[ 2(n+1)+1] ……. (1) which on simplification gives (0.6+0.3d )/(2n+1) = (0.62+0.3d)/(2n+3) Cross multiplication, componenedo dividendo (i think) etc can lead us to 2/(60+30d) = 2/(2n+1). Equating denominators, we get, 60+30d = 2n+1 or 30d = 2n+1 – 60. Now 30d is a +ve quantity . The lowest value of n for which this is true is n=30 , which gives us 30d = 1 or d = 1/30 = 0.0333 m or 3.33 cm The required answer is thus c As a further check, if we plug n = 30 in eqn (1), we get L.H.S. = R.H.S = 0.03999 
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