0
$\begingroup$

Ok, I am doing a lab and do not understand my results. We measured by hand the length in m of a tube at which our tone generator hit a point of inflection, meaning we had found the 1st then the 3rd harmonic.

We did this with 5 different frequencies. We were then told to subtract 1st-3rd to get a third data set. When graphed, this data had to have a slope of speed of sound. The third subtracted data set had least percentage error.

I do not understand WHY this is other than human error. The prompt exactly:

Why is this set of data more accurate than the others? (the third subtracted data set is set C)

Assuming you had your best results with graph c), how far off is is the first harmonic data from the ¼ wavelength indicated by c)?
How far off is the 3rd harmonic data from the ¼ wavelength indicated by c)?

Here is all the data. The order of accuracy went highest at subtracted values to lowest at 1st harmonic. I would think this would be the derived 2nd harmonic, but an open closed pipe has only odd harmonics.

enter image description here

How can I explain this?

$\endgroup$
2
$\begingroup$

Here's my understanding of your procedure: for a series of frequencies, initially 505 Hz, you adjusted the length of some sort of tube open at one end and closed at the other. You recored the first length at which you got loud resonance (15.5 cm in the first trial) and they you increased the length of the tube until you heard the second loud resonance (0.495 cm). You assumed that these lengths represented $\frac{1}{4} \lambda$ and $\frac{3}{4} \lambda$. Based on those assumptions you calculated a value of $\lambda$ for each length and then multiplied them by $f$ to get $v$ values. Then you found the difference between the two lengths, assumed that was $\frac{1}{2} \lambda$ and found the speed again.

You would think that all three measurements would lead to the same value for $\lambda$ and thus $v$ but they don't.

The piece you are missing is that the resonating air column is slightly longer than the length of the tube by a factor the depends on the diameter of the tube among other things. (See End Correction).

Rather than try to calculate exactly what this end correction factor is for your tube the easiest to take it into account is subtract two successive resonant lengths. Say $x$ is the correction factor and $l_1$ and $l_2$ are the measured lengths. The actual resonant lengths would be $l_1+x$ and $l_2+x$. When you find $\lambda$ by using: $$\frac{\lambda}{2}=l_2-l_1$$ It works because it gives the same result as: $$\frac{\lambda}{2}=(l_2+x)-(l_1+x)$$

$\endgroup$
1
$\begingroup$

The effective length of the pipe is not exactly the same as the measured length, because the air just outside the end of the pipe also vibrates.

I can't follow exactly what your spreadsheet does, since the picture doesn't show the formulas in it, but subtracting the two results will effectively cancel out this "end correction". Note that the end correction is not easy to caluculate theoretically. It mainly depends on the diameter of the pipe, but also on anything that interferes with the air flow around the open end - for example, whatever you are using to excite the air vibrations in the pipe! You can demonstrate that by partially "shading" the open end with a sheet of paper held at different distances from the end of the pipe (at a distance up to one or two times the pipe diameter), which will change the resonant frequency.

See http://isjos.org/JoP/vol5iss1/Papers/JoPv5i1-1EndCorrection.pdf, or google for "pipe end correction". (Note, the referenced paper ignores the fact that the experimenter's head will affect the results when "placing an ear near the end of the pipe", which probably accounts for the difference between the experiment and the theoretical results they quote).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.