Regarding calculation of sound wavelenth

[angeeya]

We're doing a lab to determine the ratio of specific heat of nitrogen gas (r) by using the equation v=ROOT(rRT/M), where v is the speed of sound in pure nitrogen gas, R being 8.314J/molK, T being 298K (room temp), M of nitrogen being 0.028kg/mol.

In the experiment, the function generator was connected to the speaker, whereas the oscilloscope was connected with the microphone inside the gas tube. when the microphone was moved away from the speaker, the graph displayed on the oscilloscope (voltage vs. time) only changes in amplitude (thus the voltage), but the nodes stay where they are. the pressure in the gas tube was 1.7 kpa, the frequency of wave generated by function generator was 2 kHz.

I understand that we need to determine speed of sound by v=f(lambda), frequency could be determined from f=1/period, but what about lambda? is the calculation of lambda related with the d between the microphone and the speaker? if so, how?

 

[Stonebridge]

If the waves inside the tube are standing waves, then the distance between two adjacent nodes (or antinodes) is half a wavelength.
Why do you need to determine frequency from 1/period?

 

[angeeya]

but i thought the lambda is supposed to be in the unit of "meters". since the graph is "voltage-time" graph, wouldn't that makes the wavelength in the unit of "seconds"?

actually there is no need to calculate the value of frequency, since the function generator generates the wave frequency of 2 kHz, am i right to think that?

 

[Stonebridge]

When you said "the nodes stay where they are", did you mean the nodes of the standing sound wave, or the peaks of the oscilloscope wave?
I think I have incorrectly assumed you meant the former. Sorry.
As you move the microphone you will detect the nodes (voltage max) and antinodes (voltage min) of the standing sound waves. The distance between these positions (node to node or antinode to antinode) is a half wavelength.
The generator set at 2kHz gives you the frequency.

 

[angeeya]

wow thank you very much! I really appreciate your reply!