Does dividing maximum frequency by sound frequency give accurate velocity?

[Chenkel]

TL;DR

I am trying to understand how to derive velocity of transmitter when transmitter is rotating in a circle.

Hello everyone!

I'm watching a Walter Lewin lecture and it seems to me at least that he is dividing maximum frequency of the sound by sound frequency of the transmitter to derive velocity of the transmitter, does this work? It seems that quantity would be dimensionless and velocity obviously has dimensions.

The part of the video I'm referring to is 4:50, this link requires that you queue it up, and I link the video below...


Let me know what you guys think, thank you!

 

[Frabjous]

I think the velocity should be normalized by the sound speed.

 

[Chenkel]

I think the velocity should be normalized by the sound speed.

Do you think Walter Lewin may have forgot to do that?

 

[Frabjous]

He was trying to motivate things and was not being rigorous.

 

[Chenkel]

He was trying to motivate things and was not being rigorous.

Do you know what the proper equation is?

 

[Frabjous]

Assuming that the velocity is small compared to the sound speed and that you are far away from the circle, f’/f=1+v/c
He is more rigorous in this lecture

which uses a more general equation

You should try to derive my equation from the video equation.

 

[Chenkel]

Assuming that the velocity is small compared to the sound speed and that you are far away from the circle, f’/f=1+v/c
He is more rigorous in this lecture

which uses a more general equation

You should try to derive my equation from the video equation.

Thank you for the response, I will take a look at the video!

 

[Chenkel]

I notice in the video that I linked he talks about the situation where $v\cos\theta$ is the component of velocity of the source in direction of the observer, and he writes $f' = f(1 + \frac v c \cos\theta)$, which I understand, frequency increases when it's coming toward you, and decreases when it's going away, he then writes $f' = \frac c{\lambda'}$ and $f = \frac c \lambda$ which also seems to make sense, because f is a function of signal speed and wavelength. The part I don't understand is how he uses this, and the "Taylor Series" to show the following, he writes $\lambda' = \lambda(1-\frac v c \cos \theta)$, how does this minus sign come into play? He says it works when $\frac v c$ to be much less than 1, but I don't see how this works.. I solved for the equation and I came up with $\lambda' = \lambda(1+\frac v c \cos\theta)$ why isn't this correct?

Let me know what you think, thank you.

 

[Chenkel]

I'm not sure why my equations are not rendering... Anyone else getting this?

 

[Chenkel]

I'm not sure why my equations are not rendering... Anyone else getting this?

Seems equations are rendering correctly for me now, strange, seems latex only intermittently works.

 

[Frabjous]

f’=f(1+vcosθ/c)becomes λ=λ’(1+vcosθ/c) by substitution
λ‘=λ/(1+vcosθ/c) by rearrangement
λ‘=λ(1-vcosθ/c) by Taylor expansion

 

[Ibix]

Seems equations are rendering correctly for me now, strange, seems latex only intermittently works.

It's a known bug - if there isn't already LaTeX on the page then new LaTeX doesn't render. You can refresh the page and it should appear. If you want to preview your LaTeX and it isn't rendering, copy your post to clipboard (just in case), go into preview mode, then refresh the page. The MathJax extension should have loaded and you should now be able to preview LaTeX.

 

[Chenkel]

f’=f(1+vcosθ/c)becomes λ=λ’(1+vcosθ/c) by substitution
λ‘=λ/(1+vcosθ/c) by rearrangement
λ‘=λ(1-vcosθ/c) by Taylor expansion

I read about the Taylor series today while being quite new to it, and it seems it's a approximation for a function near a certain point, and exact when an infinite number of terms are taken. I found this series for cosine, and lost all cosine terms to form something that looks a little like a geometric series (I'm not sure if it is or not.) I wonder what Walter Lewin did with the Taylor series to get the result he got.

 

[Frabjous]

Taylor (Maclaurin) series for |x|<1
(1-x)^-1=1+x+x²+x³+…
You should learn about this in calculus

 

[Chenkel]

I notice in the video that I linked he talks about the situation where $v\cos\theta$ is the component of velocity of the source in direction of the observer, and he writes $f' = f(1 + \frac v c \cos\theta)$, which I understand, frequency increases when it's coming toward you, and decreases when it's going away, he then writes $f' = \frac c{\lambda'}$ and $f = \frac c \lambda$ which also seems to make sense, because f is a function of signal speed and wavelength. The part I don't understand is how he uses this, and the "Taylor Series" to show the following, he writes $\lambda' = \lambda(1-\frac v c \cos \theta)$, how does this minus sign come into play? He says it works when $\frac v c$ to be much less than 1, but I don't see how this works.. I solved for the equation and I came up with $\lambda' = \lambda(1+\frac v c \cos\theta)$ why isn't this correct?

Let me know what you think, thank you.

I should have wrote $\lambda' =\frac \lambda {(1+\frac v c \cos \theta)}$

 

[Chenkel]

Taylor (Maclaurin) series for |x|<1
(1-x)^-1=1+x+x²+x³+…
You should learn about this in calculus

For a=0 and n=1 I got $\frac 1{(1 + x)} \approx (1 - x)$ I wonder how good of an approximation this for $|x| < 1$, as x gets close to minus one, the series approximates a very large function. I looked up the "Lagrange error bound," and it seems to be a little difficult to make sense of the error bounds for this function. Is there any easy way to determine how accurate this approximation is?

Thank you!

 

[Frabjous]

It’s a horrible approximation (edit: near -1). You can calculate the original function, so determining the error is trivial.

 

[Chenkel]

It’s a horrible approximation. You can calculate the original function, so determining the error is trivial.

I calculated the error based on your suggestion by writing the remainder and dividing by the original function $r = \frac 1{1+x}-(1-x) = \frac {x^2}{1+x}$ then the error is $e = {\frac {x^2}{1+x}} {\frac {1 + x} 1} = x^2$ so the error increases exponentially, one thing to note is that for fractional values of x it gives fairly good approximation, for example, if we let x=.1 the error is only 1 percent.

Thanks again