# Rearrange formula to show a linear relationship between Frequency and Volume

• qq194
qq194
Homework Statement
Hello,
The formula:
is to result in a linear relationship between a function of frequency f and a function of volume V.
Relevant Equations
my idea would be f² and 1/V is proportional, but the graph should result in a straight line
$$f=\frac{cs}{2\pi}\cdot\sqrt{\frac{A}{VL}}$$

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Please describe what you are trying to do and what kind of help you need. Neither the "Homework Statement" nor the pdf that you posted are very informative. It sounds like you are linearizing the equation so the graph is expected to be a straight line. What's your problem?

Also, please avoid posting pdf files. Include everything in one post. Click on the link "LaTeX Guide" (lower left) to learn how to properly render LaTeX here. Thanks.

kuruman said:
Please describe what you are trying to do and what kind of help you need. Neither the "Homework Statement" nor the pdf that you posted are very informative. It sounds like you are linearizing the equation so the graph is expected to be a straight line. What's your problem?

Also, please avoid posting pdf files. Include everything in one post. Click on the link "LaTeX Guide" (lower left) to learn how to properly render LaTeX here. Thanks.
the problem that I have is to get the function as I have described into a linear relationship between a function of frequency f and a function of volume V

qq194 said:
the problem that I have is to get the function as I have described into a linear relationship between a function of frequency f and a function of volume V
Thank you for the clarifications.
So square both sides of your equation. What do you get?

kuruman said:
Thank you for the clarifications.
So square both sides of your equation. What do you get?
i will show you my steps:
\begin{align*}
f &= \frac{c_s}{2\pi} \cdot \sqrt{\frac{A}{VL}} \\
f^2 &= \left(\frac{c_s}{2\pi}\right)^2 \cdot \frac{A}{VL} \\
f^2 \cdot \frac{VL}{A} &= \left(\frac{c_s}{2\pi}\right)^2 \\
f^2 \cdot \frac{L}{A} &= \frac{1}{V} \cdot \left(\frac{c_s}{2\pi}\right)^2 \\
\end{align*}

define c, s, A, V, L. are these independent or are there relationships between them?

gmax137 said:
define c, s, A, V, L. are these independent or are there relationships between them?
cs = speed of sound
L= length= const.
A= area = const.
V= volume = changes

qq194 said:
i will show you my steps:
\begin{align*}
f &= \frac{c_s}{2\pi} \cdot \sqrt{\frac{A}{VL}} \\
f^2 &= \left(\frac{c_s}{2\pi}\right)^2 \cdot \frac{A}{VL} \\
f^2 \cdot \frac{VL}{A} &= \left(\frac{c_s}{2\pi}\right)^2 \\
f^2 \cdot \frac{L}{A} &= \frac{1}{V} \cdot \left(\frac{c_s}{2\pi}\right)^2 \\
\end{align*}
OK. The equation for a straight line is $$y=mx+b$$where
##y =## the dependent variable
##x =## the dependent variable
##m =## the slope
##b =## the y intercept.

I assume you are making a plot of experimental results. What quantities in your experiment are the dependent and independent variables?

Switch to log scales on both axes.

PhDeezNutz, DaveE and PeroK
$$f = \frac{c_s}{2\pi} \cdot \sqrt{\frac{A}{VL}}$$
$$f^2 = \left(\frac{c_s}{2\pi}\right)^2 \cdot \frac{A}{VL}$$
$$\frac{1}{V} = \frac{2\pi f^2 \cdot L}{A \cdot c_s^2}$$
$$\ln\left(\frac{1}{V}\right) = \ln\left(2\pi f^2 \cdot L\right) - \ln\left(A \cdot c_s^2\right)$$
Chestermiller said:
Switch to log scales on both axes.

qq194 said:
$$f = \frac{c_s}{2\pi} \cdot \sqrt{\frac{A}{VL}}$$
$$f^2 = \left(\frac{c_s}{2\pi}\right)^2 \cdot \frac{A}{VL}$$
$$\frac{1}{V} = \frac{2\pi f^2 \cdot L}{A \cdot c_s^2}$$
$$\ln\left(\frac{1}{V}\right) = \ln\left(2\pi f^2 \cdot L\right) - \ln\left(A \cdot c_s^2\right)$$
or is this one better:
$$f^2 = \left(\frac{c_s}{2\pi}\right)^2 \cdot \frac{A}{VL}$$
$$\ln(f^2) = 2\ln\left(\frac{c_s}{2\pi}\right) + \ln\left(\frac{A}{VL}\right)$$
$$\ln\left(\frac{1}{V}\right) = \ln(f^2) - 2\ln\left(\frac{c_s}{2\pi}\right) - \ln\left(\frac{A}{L}\right)$$
$$y = \ln\left(\frac{1}{V}\right)$$
$$x = \ln(f^2)$$
$$m = 1$$
$$n = -2\ln\left(\frac{c_s}{2\pi}\right) - \ln\left(\frac{A}{L}\right)$$
$$y = mx + n$$

qq194 said:
$$f = \frac{c_s}{2\pi} \cdot \sqrt{\frac{A}{VL}}$$
$$f^2 = \left(\frac{c_s}{2\pi}\right)^2 \cdot \frac{A}{VL}$$
$$\frac{1}{V} = \frac{2\pi f^2 \cdot L}{A \cdot c_s^2}$$
$$\ln\left(\frac{1}{V}\right) = \ln\left(2\pi f^2 \cdot L\right) - \ln\left(A \cdot c_s^2\right)$$
$$\ln{f}=C-\frac{1}{2}\ln{V}$$where C is a constant equal to:
$$C=\ln{\left(\frac{c_s}{2\pi}\sqrt{A/L}\right)}$$So a graph of f vs V on a log-log plot will be a straight line with a slope of -1/2.

qq194 and PeroK
I wouldn't bother with logs. When you square both sides, you get
$$f^2 = \left(\frac{c_s}{2\pi}\right)^2 \frac{A}{VL}$$
Let ##y=f^2## and ##x=\dfrac{1}{V}## presumed to be the dependent and independent variables respectively. I assume that this is an experiment where you varied ##V## and measured ##f##. Presumably ##A## and ##L## are measured constants and you want to determine the speed of sound ##c_s.##

If you plot ##f^2## vs. ##\dfrac{1}{V}##, the equation should be a straight line that passes through the origin and has the form ##y=mx## where the slope ##m## is $$m=\left(\frac{c_s}{2\pi}\right)^2 \frac{A}{L}.$$You can do linear regression and find a value for ##m## which you can use to extract a value for ##c_s.## This would be harder to do if you use logarithms.

PeroK
kuruman said:
I wouldn't bother with logs. When you square both sides, you get
$$f^2 = \left(\frac{c_s}{2\pi}\right)^2 \frac{A}{VL}$$
Let ##y=f^2## and ##x=\dfrac{1}{V}## presumed to be the dependent and independent variables respectively. I assume that this is an experiment where you varied ##V## and measured ##f##. Presumably ##A## and ##L## are measured constants and you want to determine the speed of sound ##c_s.##

If you plot ##f^2## vs. ##\dfrac{1}{V}##, the equation should be a straight line that passes through the origin and has the form ##y=mx## where the slope ##m## is $$m=\left(\frac{c_s}{2\pi}\right)^2 \frac{A}{L}.$$You can do linear regression and find a value for ##m## which you can use to extract a value for ##c_s.## This would be harder to do if you use logarithms.
You don't actually have to calculate logs. All you need to do is use your plotting software to plot f vs V using default arithmetic scale. Then you double click on each axis, and in the dropdown "scales" menu, switch from linear to log. It's as easy as that. You can even have your plotting software curve fit f vs V automatically using the ##y=kx^n## "power" option; it does this transparently using regression analysis. So,, with a few clicks in the plotting software, you can accomplish everything you are suggesting with no effort on your part.

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Chestermiller said:
You don't actually have to calculate logs. All you need to do is use your plotting software to plot f vs V using default arithmetic scale. Then you double click on each axis, and in the dropdown "scales" menu, switch from linear to log. It's as easy as that. You can even have your plotting software curve fit f vs V automatically using the ##y=kx^n## "power" option; it does this transparently using regression analysis. So,, with a few clicks in the plotting software, you can accomplish everything you are suggesting with no effort on your part.
Yes, that would work in the age of spreadsheets on laptops. I (not so fondly) remember performing such tasks by creating table entries with a slide rule and plotting by hand on graph paper with a well-sharpened #2 pencil.

PeroK and topsquark
kuruman said:
Yes, that would work in the age of spreadsheets on laptops. I (not so fondly) remember performing such tasks by creating table entries with a slide rule and plotting by hand on graph paper with a well-sharpened #2 pencil.
Not quite that far back, but I did most of my work in HS on my trusty Apple II. I always generated a lot of bad data for some reason, so I got good at programming linear regression and making graphs. This was pre-Lotus, even, so I programmed my own graphs. My data sucked, but my lab write ups were pretty decent looking! (Compared to the drawn graphs, anyway.)

(There's a reason I'm a theorist, folks!)

To touch base with this thread, I agree with kuruman: I don't see why the logs would be necessary. The regression analysis of y = mx is in Excel (or LibreOffice, which is what I use).

-Dan

kuruman said:
I wouldn't bother with logs. When you square both sides, you get
$$f^2 = \left(\frac{c_s}{2\pi}\right)^2 \frac{A}{VL}$$
Let ##y=f^2## and ##x=\dfrac{1}{V}## presumed to be the dependent and independent variables respectively. I assume that this is an experiment where you varied ##V## and measured ##f##. Presumably ##A## and ##L## are measured constants and you want to determine the speed of sound ##c_s.##

If you plot ##f^2## vs. ##\dfrac{1}{V}##, the equation should be a straight line that passes through the origin and has the form ##y=mx## where the slope ##m## is $$m=\left(\frac{c_s}{2\pi}\right)^2 \frac{A}{L}.$$You can do linear regression and find a value for ##m## which you can use to extract a value for ##c_s.## This would be harder to do if you use logarithms.

Okay that means
kuruman said:
I wouldn't bother with logs. When you square both sides, you get
$$f^2 = \left(\frac{c_s}{2\pi}\right)^2 \frac{A}{VL}$$
Let ##y=f^2## and ##x=\dfrac{1}{V}## presumed to be the dependent and independent variables respectively. I assume that this is an experiment where you varied ##V## and measured ##f##. Presumably ##A## and ##L## are measured constants and you want to determine the speed of sound ##c_s.##

If you plot ##f^2## vs. ##\dfrac{1}{V}##, the equation should be a straight line that passes through the origin and has the form ##y=mx## where the slope ##m## is $$m=\left(\frac{c_s}{2\pi}\right)^2 \frac{A}{L}.$$You can do linear regression and find a value for ##m## which you can use to extract a value for ##c_s.## This would be harder to do if you use logarithms.

okay thanks, but the rearranged equation that describes the linear relationship between a function of frequency f and a function of volume is f² and 1/V?because I have to rearrange the equation to get this result

kuruman said:
Yes, that would work in the age of spreadsheets on laptops. I (not so fondly) remember performing such tasks by creating table entries with a slide rule and plotting by hand on graph paper with a well-sharpened #2 pencil.
Even back in 1963 when I graduated from college, before spreadsheets on laptops, K&E was selling log-log graph paper with various numbers of decades on each axis; I would have known to plot f vs V on log-log paper.

nasu and Lord Jestocost
Chestermiller said:
$$\ln{f}=C-\frac{1}{2}\ln{V}$$where C is a constant equal to:
$$C=\ln{\left(\frac{c_s}{2\pi}\sqrt{A/L}\right)}$$So a graph of f vs V on a log-log plot will be a straight line with a slope of -
Thank you very much! I tried their equation and it worked! The graph looks good too. What do you call the x and y axis? And can you also use the logarithms to form a linear relationship between a function of frequency f and a function of volume V?or is the equation you gave me already the linear relationship?
Greetings

qq194 said:
Thank you very much! I tried their equation and it worked! The graph looks good too. What do you call the x and y axis? And can you also use the logarithms to form a linear relationship between a function of frequency f and a function of volume V?or is the equation you gave me already the linear relationship?
Greetings
You label the x and y axes F and V. The equation I gave is already the linear relationship.

qq194
Chestermiller said:
You label the x and y axes F and V. The equation I gave is already the linear relationship.
I have tried to show the derivation of your equation, can you help me if there is an error in the derivation?or do you have a better derivation how you come to the equation? Because I like your equation very much and I would like to try to understand the derivation

$$f = \frac{{cs}}{{2\pi}} \sqrt{\frac{A}{VL}}$$
$$2\pi f = cs \sqrt{\frac{A}{VL}}$$
$$(2\pi f)^2 \cdot VL = (cs)^2 \cdot A$$
$$V = \frac{{(cs)^2 \cdot A}}{{(2\pi f)^2 \cdot L}}$$
$$\ln V = \ln \left( \frac{{(cs)^2 \cdot A}}{{(2\pi f)^2 \cdot L}} \right)$$
$$\ln V = 2 \ln(cs) + \ln A - 2 \ln(2\pi f) - \ln L$$
$$C = \ln\left(\frac{{cs}}{{2\pi}} \sqrt{\frac{A}{L}}\right) = \ln(cs) - \ln(2\pi) + \frac{1}{2} \ln\left(\frac{A}{L}\right)$$
$$\ln V = C + \ln(cs) - \ln(2\pi) - \ln f - \ln L$$
$$\ln f = C - \frac{1}{2} \ln V$$

No. $$f=\left(\frac{cs}{2\pi}\right)\left(\frac{A}{VL}\right)^{\frac{1}{2}}$$
$$\log{f}=\log{\left(\frac{cs}{2\pi}\right)}+\log{\left(\frac{A}{VL}\right)^{\frac{1}{2}}}$$
$$\log{\left(\frac{A}{VL}\right)^{\frac{1}{2}}}=\log{\sqrt{\frac{A}{L}}}-\frac{1}{2}\log{V}$$

qq194
qq194 said:
Okay that means

okay thanks, but the rearranged equation that describes the linear relationship between a function of frequency f and a function of volume is f² and 1/V?because I have to rearrange the equation to get this result
Except for squaring both sides, you don't have to rearrange anything. If you let ##y=f^2## and ##x=1/V##, you get a straight line ##y=mx## that you can fit to get a value for ##m##. Then you will have to solve the equation $$\left(\dfrac{c_s}{2\pi}\right)^2\dfrac{A}{L}=m$$ for ##c_s##. You can do this because you already know the power dependence of ##f## on ##V##, namely that the frequency is proportional to the inverse square root of the volume. All of this can be done on paper.

The method suggested by @Chestermiller is very useful when you are looking for the power law between two quantities related by ##y=(\text{const})x^n##. Constructing a log-log plot gives you a straight line. Then the slope of the line ## \Delta y/\Delta x##, is the power dependence of ##y## on ##x##, here ##n=-\frac{1}{2}.## That much can be done on paper. However, determining the value of ##(\text{const})## from the plot could be tricky and is best left to a fitting routine.

The moral is don't do anything on paper if you don't have to.

qq194
Chestermiller said:
No. $$f=\left(\frac{cs}{2\pi}\right)\left(\frac{A}{VL}\right)^{\frac{1}{2}}$$
$$\log{f}=\log{\left(\frac{cs}{2\pi}\right)}+\log{\left(\frac{A}{VL}\right)^{\frac{1}{2}}}$$
$$\log{\left(\frac{A}{VL}\right)^{\frac{1}{2}}}=\log{\sqrt{\frac{A}{L}}}-\frac{1}{2}\log{V}$$
I have entered the normal V and f values in the diagram, does the slope have to be -1/2 when running a regression leanest through the points as in your equation. And do I have to adjust the values through the equation when I enter them

qq194 said:
I have entered the normal V and f values in the diagram, does the slope have to be -1/2 when running a regression leanest through the points as in your equation. And do I have to adjust the values through the equation when I enter them
The slope is what the fitting algorithm says it is. If you expect -1/2 and it is not (within experimental error), then you have to interpret why.

I do not understand what you mean by "adjust" the values. Adjust them how and to what purpose?

Let’s see the data.

gmax137
Chestermiller said:
Let’s see the data.
Yes I agree! I know that the OP is really about working with the experimental data but I am curious as to what is varied (apparently, "V") and what is measured (apparently "f" but what is "f").

gmax137 said:
Yes I agree! I know that the OP is really about working with the experimental data but I am curious as to what is varied (apparently, "V") and what is measured (apparently "f" but what is "f").
"f" is frequency "V" is some kind of volume and ##c_s## is the speed of sound. By dimensional analysis, ##A## and ##L## are, respectively, some kind of area and length. I know all this because initially OP attached a pdf which was since deleted. There was no information on the physical situation in the pdf.

gmax137
kuruman said:
"f" is frequency "V" is some kind of volume and ##c_s## is the speed of sound. By dimensional analysis, ##A## and ##L## are, respectively, some kind of area and length. I know all this because initially OP attached a pdf which was since deleted. There was no information on the physical situation in the pdf.
There are different volume bet and accordingly different frequencies. I have recorded 7 values, so now I have 7 volume values and their corresponding frequency. @Chestermiller gave me a nice Ln equation, I put in his equation the values, you can see the linear relationship. Do I have to plot the normal values V and f without having put them into @Chestermiller equation. Or do I have to draw the logarithm values I have from the equation into a log diagram.I knew the equation and the graph as @kuruman did it before, and it works, but I would like to try the logarithm variant as well

qq194 said:
There are different volume bet and accordingly different frequencies. I have recorded 7 values, so now I have 7 volume values and their corresponding frequency. @Chestermiller gave me a nice Ln equation, I put in his equation the values, you can see the linear relationship. Do I have to plot the normal values V and f without having put them into @Chestermiller equation. Or do I have to draw the logarithm values I have from the equation into a log diagram.I knew the equation and the graph as @kuruman did it before, and it works, but I would like to try the logarithm variant as well
please provide the f vs V data, and I will show you what I mean. Plus, I have already answered these questions in a previous post.

Chestermiller said:
please provide the f vs V data, and I will show you what I mean. Plus, I have already answered these questions in a previous post.
V= 0,000295m^3 f=172Hz
V= 0,000245m^3. f=188Hz
V=0,000195m^3 f=211Hz
V= 0,000145m^3 f=245Hz
V=0,000095m^3 f= 303 Hz
V=0,000045m^3 f= 440Hz

qq194 said:
V= 0,000295m^3 f=172Hz
V= 0,000245m^3. f=188Hz
V=0,000195m^3 f=211Hz
V= 0,000145m^3 f=245Hz
V=0,000095m^3 f= 303 Hz
V=0,000045m^3 f= 440Hz

Chestermiller said:
now it works, thank you , with me the graph looked like a straight line, only my associated function equation y=mx+b was not as perfect as yours. Many thanks, what program did you use?

qq194 said:
now it works, thank you , with me the graph looked like a straight line, only my associated function equation y=mx+b was not as perfect as yours. Many thanks, what program did you use?
Kaleidagraph

Chestermiller said:
Kaleidagraph
Chestermiller said:
Chestermiller said:
Kaleidagraph
thank you so much for your really great support!

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