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I'm studying high school physics and I've encountered this question:

An air column that is open at both ends has a distance of 24.0 cm from one resonant length to another. What is the wavelength of sound that is in resonance with this tube? How would the wavelength be affected if the tube was closed at one end?

I solved the first part of the question by noticing that the distance between resonant lengths is equal to the inter-nodal distance $d_n=\frac{1}{2}\lambda$. So $\lambda=(2) (0.24m)=0.48m$.

The second part, however, confuses me. I tested in real life by blowing in a paper tube and I found that frequency (pitch) is lower when the end is closed compared to when it is open. So, by the wave equation $v=\lambda f$, assuming the speed of sound is constant, wavelength $\lambda$ must increase to compensate for the frequency $f$ decreasing. I don't see how and why $\lambda$ must increase.

Am I doing or understanding something wrong?

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the shortest resonating wave in a tube with one end closed is $\lamba/4$ a knot at the closed end and a maximum at the open end, so the tubelenght is $\lamba/4$ while in the open tube it is $\lamba/2$ but both have to be 24cm longer for the next occurring resonance

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  • $\begingroup$ I don't quite understand your answer. It seems you have just repeated the fact that "An air column that is open at both ends has a distance of 24.0 cm from one resonant length to another." $\endgroup$ – Sean Xie Apr 7 '20 at 0:38
  • $\begingroup$ I tried to say that the distance between two resonances is the same for the open and closed tube, but the closed tube has a smaller length . for the first resonance. the resonancee length of a closed tube are $\lambda/4+n*\lambda/2$ for the open tube it is $\lambda/2+n*\lambda/2$ $\endgroup$ – trula Apr 7 '20 at 14:31
  • $\begingroup$ Ah, so frequency $f$ increases? $\endgroup$ – Sean Xie Apr 7 '20 at 14:33

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