Physics Help Forum Need some help please with resonant lengths.

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 Jun 24th 2017, 05:02 PM #1 Junior Member   Join Date: Jun 2017 Posts: 15 Need some help please with resonant lengths. Hi, I'm just beginning physics and I'm having some trouble. Even when I get the right answers they just don't feel correct or entirely complete. I'm just hoping someone can let me know if I did this right, and what I could do to add to it if necessary. I apologize for the messiness I couldn't find symbols, I hope it is clear enough. A hollow tube chime is open at both ends, the chime is 0.54m in length, the speed of sound is 346m/s. Finding the third resonant length and it's frequency: L = wavelength/2 0.54m = wavelength/2 2(0.54m) = wavelength 1.08m = wavelength third resonant length: L3 L3 = 3(wavelength)/2 L3 = 3(1.08m)/2 L3 =1.62m third resonant length: 1.62m f = v/wavelength = 346/1.62 = 213.58 Hz frequency of third resonant length: 213.58 Hz Thanks very much for your help.
 Jun 24th 2017, 08:04 PM #2 Senior Member     Join Date: Aug 2008 Posts: 113 for an open-ended tube of length $L$ ... $f_1 \implies L = \dfrac{\lambda}{2} \implies \lambda = 2L$ $f_2 \implies L = \lambda$ $f_3 \implies L = \dfrac{3\lambda}{2} \implies \lambda = \dfrac{2L}{3}$ $f_4 \implies L = 2\lambda \implies \lambda = \dfrac{L}{2}$ Note the sequence for values of $\lambda$ ... $\dfrac{2L}{1}, \, \dfrac{2L}{2}, \, \dfrac{2L}{3}, \, \dfrac{2L}{4}, \, ...$
 Jun 24th 2017, 08:47 PM #3 Junior Member   Join Date: Jun 2017 Posts: 15 I originally didn't see how that helped me at all, but I see now what I did wrong. Thank you for your help.

 Tags frequency, hertz, lengths, resonant, wave

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