As the title suggests, I'm wondering how I handled the following exercise checks out:

A loudspeaker is placed close to one end of a pipe that is open at both ends. The loudspeaker is driven by a signal generator of adjustable frequency. At $660 Hz$ a maximum in the sound volume (a resonance) is heard. The frequency of the signal generator is slowly decreased, and it is noted that the next frequency at which a maximum is heard is $550 Hz$. The speed of sound in air is $343 \ ms^{−1}$. Determine the length of the pipe and the lowest frequency at which the air column in it will resonate, explaining your reasoning.

Okay, so since this is an open pipe, it implies that at $x=0$ and $x=L$, we require an antinode. The equation for a standing sound wave has an argument of $cos(k_n \ x)$ or $sin(k_n \ x)$ for its amplitude, depending on the boundary conditions.

In this case, I've elected to state:

$k_n \ L = n \pi$

Where $n$ is an integer. This is because we require a cosine amplitude argument in order to satisfy both $x = 0 \implies s \ne 0$ and $x = L \implies s \ne 0$ given the fact that the pipe is open at both ends. $s$ is the displacement of particles within the medium from equilibrium, parallel to wave propogation velocity.

From this being fulfilled, I believe I am justified to argue this.

Now, since $660 \ Hz$ was a normal mode frequency (will define with $n_1$), and then was turned down to the next normal mode frequency of $n_2$, this means that $n_1$ was the next integer mode after $n_2$.

$$\implies n_1 = n_2 + 1$$

From this, I argue, due to the condition that the two frequencies given are resonant frequencies:

$$k_{n_1} \approx 12.1 \ m^{-1}$$

$$\implies L = \frac{n_1 \pi}{12.1} = \frac{(n_2+1)\pi}{12.1}$$

And from the second resonant frequency..

$$k_{n_2} \approx 10.07 \ m^{-1}$$

$$\implies L = \frac{n_2 \pi}{10.07}$$

$$\frac{n_2 \pi}{10.07} = \frac{(n_2+1)\pi}{12.1}$$

$$\implies n_2 = 5$$

From this, $L$ can be found, and the lowest resonant frequency can be found by the relation:

$$n = \frac{k_n \ L}{\pi}$$

With $n = 1$.

Are my arguments fair?

  • $\begingroup$ There is some superfluous work I think. Since you know that each frequency is a multiple of the fundamental, all you need to solve is $\frac{550}{n}=\frac{660}{n+1}=f_0$. This is to get that you have have 5th and 6th modes. $\endgroup$ Commented Mar 9, 2018 at 18:57

1 Answer 1


Here are some thoughts:

  • The standing wave pattern for any set of boundary conditions may be written as the sum of cosine and sine waves. The boundary conditions determine what form $k_nL$ takes. For your case of an open-open tube the acoustic pressure is zero at the boundaries (i.e., the pressure is equal to the ambient pressure). Thus, at $x=0,\ L$ you require $p=0$ and from these conditions you obtain the requirement that $k_nL=n\pi$. All of this is to say that you can't "elect" to have this condition. You should also show your work.
  • The quantity $k_n$ is not a frequency. You should report $f_n=k_nc_0/2\pi$, where $c_0$ is the sound speed.
  • You should show your work for how you obtained the values 12.1 m$^{-1}$ and 10.07 m$^{-1}$.
  • $\begingroup$ The OP doesn't assume that the two given frequencies are the two lowest modes and doesn't say that the k values are frequencies. The work shows that they got the 5th and 6th modes actually. $\endgroup$ Commented Mar 9, 2018 at 18:39
  • $\begingroup$ Right. That is why $m$ is still to be determined. If they were the lowest two frequencies then $m$ would equal 1. $\endgroup$
    – Michael M
    Commented Mar 9, 2018 at 18:41
  • $\begingroup$ Yeah in their work m is 5 with how you are defining m. I'm wondering why you are telling them to correct things that are already right. $\endgroup$ Commented Mar 9, 2018 at 18:44
  • $\begingroup$ I see it now. I got distracted by their definitions of $n_1$ and $n_2$. I have removed this comment from my answer. $\endgroup$
    – Michael M
    Commented Mar 9, 2018 at 18:46
  • $\begingroup$ The most important part of this answer is that one cannot elect the boundary conditions. The pressure is zero at $x=0$ and at $x=L$ so the pressure is describe by a sine, not a cosine, and $k_nL=n\pi$ $\endgroup$
    – Alfred
    Commented Nov 27, 2019 at 6:59

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