Area Under Frequency versus Time Curve meaning?

In summary, the conversation discussed two methods for finding the velocity of waves in a string. Method one involved plotting frequency on the y-axis and 1/wavelength on the x-axis and drawing a best fit line to determine the slope, while method two involved plotting frequency on the y-axis and wavelength on the x-axis and taking the area under the curve. The problem with method two is that it gives a nonsensical answer and does not follow the proper relationship v=λf. This is because the relationship between velocity and frequency is not represented by the area under the curve, unlike in the case of finding distance from a velocity-time curve. It is important to understand the proper relationships and use dimensional analysis to determine the correct methods for finding quantities
  • #1
GregB777
5
1
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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  • #3
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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  • #5
haruspex said:
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?
 
  • #6
GregB777 said:
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.
 
  • #7
haruspex said:
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
 
  • #8
haruspex said:
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?
 
  • #9
GregB777 said:
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##?
 

Related to Area Under Frequency versus Time Curve meaning?

What is the area under the frequency versus time curve?

The area under the frequency versus time curve is a measure of the total energy or power contained in a signal over a specific period of time. It represents the integral of the signal's power spectrum.

Why is the area under the frequency versus time curve important?

The area under the frequency versus time curve is important because it provides information about the distribution of energy or power in a signal over time. It can be used to analyze and compare different signals, and can also help identify specific frequency components within a signal.

How is the area under the frequency versus time curve calculated?

The area under the frequency versus time curve is calculated by integrating the power spectrum of a signal over a specific time interval. This can be done using mathematical tools such as calculus or by using specialized software or equipment.

What does a larger area under the frequency versus time curve indicate?

A larger area under the frequency versus time curve indicates that the signal contains more energy or power over the given time interval. This could be due to a higher amplitude or a wider distribution of frequency components within the signal.

What can cause changes in the area under the frequency versus time curve?

Changes in the area under the frequency versus time curve can be caused by various factors such as changes in the signal's amplitude, frequency components, or duration. It can also be affected by external factors such as noise or interference. Additionally, changes in the measurement or analysis methods can also impact the calculated area under the curve.

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