05-07-2011, 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. |

05-07-2011, 10:22 AM |
#2 | |

Super Moderator Join Date: Apr 2008 Location: On the dance floor, baby!
Posts: 1,921
| Quote:
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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05-07-2011, 11:40 AM |
#3 | |

Junior Member Join Date: Apr 2008
Posts: 6
| Naw, not that complicated. Quote:
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. | |

05-07-2011, 03:16 PM |
#4 | |

Super Moderator Join Date: Apr 2008 Location: On the dance floor, baby!
Posts: 1,921
| Quote:
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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05-08-2011, 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. |

05-21-2011, 10:17 AM |
#6 |

PHF Helper Join Date: Feb 2009
Posts: 1,422
| 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 =0.04Compare 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 They do not represent resonance at different frequencies. Thus we may write,but different numbers of this wavelength.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 us30d = 1 or d = 1/30 = 0.0333 m or 3.33 cm The required answer is thus cAs 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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