Area Under Frequency versus Time Curve meaning?

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Homework Statement:
Find the velocity of waves in a given string.
Relevant Equations:
frequency*wavelength=velocity
Hello: Let's say you have a string and get data by changing the frequency a transverse wave in the string to get different standing modes. You measure the wavelength of each mode for each frequency. That is, the data you get are frequency and wavelength. Now, you are trying to find the velocity of the waves in the string.

I can see two ways to do this, but the second method gives a nonsensical answer and I don't know why!

(1) Method one: Plot frequency on the y-axis and 1/wavelength on the x-axis. Draw a best fit line. The slope of the best-fit line with be the velocity since the equation is: f =v* (1/wavelength)

(2) Method two: Plot frequency on the y-axis and wavelength on the x-axis. Then take the area under the curve to find the velocity since f*wavelength=velocity.

The problem I have with method two is that if you add more data points, the area under the curve will increase! That is, the velocity will increase with more data points... and that makes no sense! The velocity is the the velocity! It should not increase if you add more data.

Can someone please explain what the area under the curve of a frequency vs. wavelength curve represents and what is wrong with the calculation in method two??

Thanks!
 

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haruspex
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An integral is a smoothed form of summation. ##W=\int F.dx## is the continuous version of ##\Sigma F_i.dx_i##.
In your experiment, the frequency and wavelength do not change continuously. You have wavelengths of ##\frac{2L}n## only. But even if they could somehow be made continuous, the discrete relationship is ##v=\lambda_i f_i##, not ##v=\Sigma f_i.d\lambda_i##.
Summing over the range ##\lambda=\frac{2L}{n_1}## to ##\lambda=\frac{2L}{n_2-1}## should produce something like ##v(\ln(n_2)-\ln(n_1))##.
 
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An integral is a smoothed form of summation. ##W=\int F.dx## is the continuous version of ##\Sigma F_i.dx_i##.
In your experiment, the frequency and wavelength do not change continuously. You wavelengths of ##\frac{2L}n## only. But even if they could somehow be made continuous, the discrete relationship is ##v=\lambda_i f_i##, not ##v=\Sigma f_i.d\lambda_i##.
Thank you for the thoughtful reply. I also thought the discrete relationship had something to do with it.

However, if that is so, why does plotting a best-fit line over discrete data in the f vs. 1/λ curve actually still work?
 
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haruspex
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why does plotting a best-fit line over discrete data in the f vs. 1/λ curve actually still work?
Because the relationship is ##v=\frac f{ 1/\lambda}## independently at each point, it makes no difference if you treat it as ##\frac{\Delta f}{\Delta(1/\lambda)}##.

Presumably you are thinking in terms of differentiation and integration being inverse processes. The slope from plotting f vs. 1/λ corresponds to differentiating f wrt 1/λ. Inverting that would mean integrating v wrt 1/λ to get f, not integrating f wrt λ to get v.
 
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Because the relationship is ##v=\frac f{ 1/\lambda}## independently at each point, it makes no difference if you treat it as ##\frac{\Delta f}{\Delta(1/\lambda)}##.

Presumably you are thinking in terms of differentiation and integration being inverse processes. The slope from plotting f vs. 1/λ corresponds to differentiating f wrt 1/λ. Inverting that would mean integrating v wrt 1/λ to get f, not integrating f wrt λ to get v.
I actually wasn't thinking about the relationship between differentiation and integration, but rather the first part of your answer. (The student that asked me this question does not know calculus.)

However, your observation is very helpful.

This question came from one of my students who didn't understand why he could take the area under a v-t curve to get a distance (since if you do dimensional analysis, (m/s)*s=meters) but COULD NOT take the area under a f-λ curve and get velocity (since if you do dimensional analysis (1/s)*(m)=m/s.).

He was taught to use dimensional analysis to determine proper relationships and does not understand why, in this case, you can't simply transfer one area-under-the-curve method to another. The answer is actually pretty subtle and is because v=λifi and NOT v=Σfii
 
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Because the relationship is ##v=\frac f{ 1/\lambda}## independently at each point, it makes no difference if you treat it as ##\frac{\Delta f}{\Delta(1/\lambda)}##.

Presumably you are thinking in terms of differentiation and integration being inverse processes. The slope from plotting f vs. 1/λ corresponds to differentiating f wrt 1/λ. Inverting that would mean integrating v wrt 1/λ to get f, not integrating f wrt λ to get v.
In general, is there a rule I can give my student to help him to decide WHEN using an area under the curve method is OK and when it isn't?

It doesn't JUST have to do with it being a continuous or non-continuous function, does it? (..because you can approximate a continuous function with more data.)

Or maybe it does?
 
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haruspex
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In general, is there a rule I can give my student to help him to decide WHEN using an area under the curve method is OK and when it isn't?

It doesn't JUST have to do with it being a continuous or non-continuous function, does it? (..because you can approximate a continuous function with more data.)

Or maybe it does?
The first thing to explain to the student is that dimensional analysis has its limitations. Occasionally I see a thread posted on this forum where a student has applied a standard equation to given data, apparently based on plugging in data values that happen to have the right dimension, regardless of any other logical relationship between what the data value represents and what the variable in the equation represents.
As I respond, a standard equation is incomplete without specifications of what the variables mean and in what context it operates.

But more specifically in this case, the distinction is whether the underlying relationship is between the instantaneous values or increments. I.e., is it ##z=yx## or ##\Delta z= y\Delta x##?
 

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