Vay Posted May 7, 2013 Share Posted May 7, 2013 (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 May 7, 2013 by Vay Link to comment Share on other sites More sharing options...

Przemyslaw.Gruchala Posted May 7, 2013 Share Posted May 7, 2013 (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 May 7, 2013 by Przemyslaw.Gruchala Link to comment Share on other sites More sharing options...

swansont Posted May 7, 2013 Share Posted May 7, 2013 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 Link to comment Share on other sites More sharing options...

elfmotat Posted May 7, 2013 Share Posted May 7, 2013 The easiest way to demonstrate this is to simply look at a graph of cos^{2}(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. Link to comment Share on other sites More sharing options...

Vay Posted May 8, 2013 Author Share Posted May 8, 2013 (edited) The easiest way to demonstrate this is to simply look at a graph of cos^{2}(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 May 8, 2013 by Vay Link to comment Share on other sites More sharing options...

swansont Posted May 8, 2013 Share Posted May 8, 2013 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/ Link to comment Share on other sites More sharing options...

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