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Cosine identity for energy transmission of waves and speed of waves.


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Posted (edited)

In reference to the average change in kinetic energy over time for waves moving along a string, (1/2)*u*v*(w^2)*(ymax^2)(average of ((cos(kx-wt))^2) over time), my textbook says "the average value of the square of a cosine function over an integer number of periods is (1/2)", thus the above equation reduces to (1/4)*u*v*(w^2)*(ymax^2). Can anyone explain the cosine identity, or whatever it is, used here(the other variables are not important, I only have question about the identity in the quotations)?

 

Also, what is the difference between the speed of waves defined as v = (wavelength)*(frequency) and v= square root of (tension/linear density)? My book says the speed is related to wavelength and frequency, but it is set by properties of the medium, which is the tension and linear density. Can someone explain this clearer?

Edited by Vay
Posted (edited)

Also, what is the difference between the speed of waves defined as v = (wavelength)*(frequency) and v= square root of (tension/linear density)? My book says the speed is related to wavelength and frequency, but it is set by properties of the medium, which is the tension and linear density. Can someone explain this clearer?

 

http://en.wikipedia.org/wiki/Refractive_index

 

v = wavelength * frequency = c - in vacuum (because refractive_index of vacuum is 1)

and

v = c / refractive_index = wavelength * frequency / refractive_index - in some medium

 

f.e. refractive index of water is 1.333

so

v = c / 1.333 = 0.75c

and also

wavelength = wavelength in vacuum / 1.333

Edited by Przemyslaw.Gruchala
Posted

http://en.wikipedia.org/wiki/Refractive_index

 

v = wavelength * frequency = c - in vacuum (because refractive_index of vacuum is 1)

and

v = c / refractive_index = wavelength * frequency / refractive_index - in some medium

 

f.e. refractive index of water is 1.333

so

v = c / 1.333 = 0.75c

and also

wavelength = wavelength in vacuum / 1.333

 

The OP clearly states "waves moving along a string" Speed of light has nothing to do with the problem

 

In reference to the average change in kinetic energy over time for waves moving along a string, (1/2)*u*v*(w^2)*(ymax^2)(average of ((cos(kx-wt))^2) over time), my textbook says "the average value of the square of a cosine function over an integer number of periods is (1/2)", thus the above equation reduces to (1/4)*u*v*(w^2)*(ymax^2). Can anyone explain the cosine identity, or whatever it is, used here(the other variables are not important, I only have question about the identity in the quotations)?

 

http://www.ditutor.com/integrals/integral_cos_squared.html

Posted

The easiest way to demonstrate this is to simply look at a graph of cos2(x). It's a periodic function, whose amplitude varies from 0 to 1. It's not unreasonable, therefore, to suspect that the average value of this function from 0 to 2pi is simply 1/2. If you do the integral, this turns out to be the case.

Posted (edited)

The easiest way to demonstrate this is to simply look at a graph of cos2(x). It's a periodic function, whose amplitude varies from 0 to 1. It's not unreasonable, therefore, to suspect that the average value of this function from 0 to 2pi is simply 1/2. If you do the integral, this turns out to be the case.

 

Isn't the intergral of (cos(x))^2, from 0 to 2pi, equal to pi/2? Or are you taking the integral of something else. The graphical interpretation makes more sense, where we're just taking the average of the two extreme y values of (cos(x))^2.

Edited by Vay
Posted

Isn't the intergral of (cos(x))^2, from 0 to 2pi, equal to pi/2? Or are you taking the integral of something else. The graphical interpretation makes more sense, where we're just taking the average of the two extreme y values of (cos(x))^2.

 

The integral isn't the average value — you were asking about the cosine identity used, which was given in the link. You have to take the next step and divide by the size of the interval.

http://archives.math.utk.edu/visual.calculus/5/average.1/

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