Kundt's Tube Experiment


Today, we have taught kundt's Tube experiment. In the this experiment, a stationary wave is set up inside a closed tube. Lycopodium powser is spread evenly inside the tube. The node positions of the stationary wave will form heaps on the powders. Distance d between two heaps is measured and the frequency of the wave is denoted as f.

It is said in our textbook that the speed of sound in air is given by c=2df. As in the experiment, we can change the frequency of the signal generator , thus changes the frequency of the wave. If I change the frequency, I know that the distance d will also change. But will they change in proportion? Will the final speed be different?

I have tried to prove this mathematically, and what I obtain is: AS f is proportional to 1/lambda when frequency f doubled, lambda prime becomes one half of lambda. And as lambda is proportional to 2d, when lambda prime becomes one half of lambda, d prime becomes 1/4 d. As v=2df, v prime = 2 (1/4 d)X(2f) = df , which is not equal to v !

Is there anything wrong in my calculations?

Also, I wonder why the tube has to be dry? is it beecause we want to avoid the powders from sticking together?


The speed of sound as measured by the Kundt's tube apparatus is 2 times the distance between nodes times the frequency of the oscillator, assuming that the oscillator puts out a single frequency only and not some more complex waveform. The speed of sound is related to the air pressure and temperature which does not change in your experiment. So when you change the frequency of the oscillator, the distance between the nodes in the tube will change so as to give you the same result for sound speed.

The old d was 1/2 the old wave length. The new d is 1/2 the new wavelength. If the wavelength was halved, then the new d is 1/2 the old d, not 1/4. The new d is 1/4 the old wavelength.

You are right about the powder being dry so it is free to respond to the small pressure difference carried by the sound wave.