Here is my approach to this:

neglecting any thermal expansion of the pipe:

By the Laplace formula for the speed of sound,

$V=\sqrt{\frac{\gamma P}{\rho}}$ where P is the pressure, $\gamma$ is the adiabatic constant and $\rho$ is the density of the medium.

Assuming the gas to be an ideal gas, we can use the ideal gas equation.
Hence we have:

$V=\sqrt{\frac{\gamma RT}{M}}$ where R is the gas constant, T is the absolute temperature, and M is the molar mass of air

So, when we increase the temperature, clearly, the velocity would increase as well. Coming to the fundamental frequency, we know

$f_0$ (fundamental frequency) $\alpha$ V (velocity of sound)

Hence, the fundamental frequency would also increase.

But how would we get the variance of the wavelength with temperature? I thought of using the relation

$V=f\lambda$ where V is the velocity of sound, f is the frequency and $\lambda$ is the wavelength.

According to which, the wavelength should also increase but according to my book, that's not right. Why does this happen?

  • $\begingroup$ so thermal expansion of the organ pipe itself is neglected? $\endgroup$
    – sleepy
    Mar 30, 2021 at 13:39
  • $\begingroup$ @sleepy yes it is. $\endgroup$ Mar 30, 2021 at 14:04

2 Answers 2


HINT: Sound waves in an organ pipe are examples of standing waves. The allowed wavelengths of a standing wave on (for example) a string are not arbitrary, but are instead determined by the length of the string.


The wavelength is independent of temperature.

enter image description here

You can see here that the wavelength depends only on the length of the organ pipe and the harmonic of the resonance, rather than the temperature of the gas itself.

The only way of changing the wavelength is by increasing the harmonic or the length of the pipe itself.

Hope this helps answer your question.


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