Sriman Dutta Posted December 9, 2016 Share Posted December 9, 2016 (edited) Hi everybody, We all know that in mathematics, any wave can be thought of as the plot of a circle in the 2D coordinate plane (considering 2D waves only). A wave [math]W[/math] may be represented as: [math]W(x,t)=Acos(kx-\omega t)[/math] Where [math]W(x,t)[/math] is the function of the wave's position [math]x[/math] and time [math]t[/math], which gives the displacement from the x-axis, [math]k[/math] is the wavenumber, [math]\omega[/math] is the angular frequency of the circle; [math]\omega=2\pi f[/math], where [math]f[/math] is the frequency of the wave. Thus the wave [math]W[/math] is a curved line, consisting of points of the form [math](x,Acos(kx-\omega t))[/math] But, I think (and that's my question too) that the position [math]x[/math] and time [math]t[/math] can be represented as functions of the angle of the circle [math]\theta[/math]. Here's how... Let the wave velocity be [math]v[/math]. Then, [math] v=\frac{x}{t}[/math] Or, [math]x=vt[/math] But, [math]v=f\lambda[/math], [math]f[/math] is frequency and [math]\lambda[/math] is the wavelength. So, [math]x=f\lambda t[/math] But, [math]f=\frac{\omega}{2\pi}[/math] So, [math]x=\frac{\omega}{2\pi}\lambda t[/math] But, [math]\omega=\frac{\theta}{t}[/math] So, [math]x=\frac{\theta \lambda}{2\pi}[/math] Similarly, [math] t=\frac{\theta}{\omega}[/math] Am I correct ? Edited December 9, 2016 by Sriman Dutta Link to comment Share on other sites More sharing options...
studiot Posted December 9, 2016 Share Posted December 9, 2016 What happened to k in your working? It is in there because ther are multiple solutions to the equation v = x/t Link to comment Share on other sites More sharing options...
Sriman Dutta Posted December 10, 2016 Author Share Posted December 10, 2016 So, will it be: v=kx/t ? Link to comment Share on other sites More sharing options...
Sriman Dutta Posted December 16, 2016 Author Share Posted December 16, 2016 Can anyone guide me? Link to comment Share on other sites More sharing options...
DevilSolution Posted December 17, 2016 Share Posted December 17, 2016 (edited) Hi everybody, We all know that in mathematics, any wave can be thought of as the plot of a circle in the 2D coordinate plane (considering 2D waves only). A wave [math]W[/math] may be represented as: [math]W(x,t)=Acos(kx-\omega t)[/math] Where [math]W(x,t)[/math] is the function of the wave's position [math]x[/math] and time [math]t[/math], which gives the displacement from the x-axis, [math]k[/math] is the wavenumber, [math]\omega[/math] is the angular frequency of the circle; [math]\omega=2\pi f[/math], where [math]f[/math] is the frequency of the wave. Thus the wave [math]W[/math] is a curved line, consisting of points of the form [math](x,Acos(kx-\omega t))[/math] But, I think (and that's my question too) that the position [math]x[/math] and time [math]t[/math] can be represented as functions of the angle of the circle [math]\theta[/math]. Here's how... Let the wave velocity be [math]v[/math]. Then, [math] v=\frac{x}{t}[/math] Or, [math]x=vt[/math] But, [math]v=f\lambda[/math], [math]f[/math] is frequency and [math]\lambda[/math] is the wavelength. So, [math]x=f\lambda t[/math] But, [math]f=\frac{\omega}{2\pi}[/math] So, [math]x=\frac{\omega}{2\pi}\lambda t[/math] But, [math]\omega=\frac{\theta}{t}[/math] So, [math]x=\frac{\theta \lambda}{2\pi}[/math] Similarly, [math] t=\frac{\theta}{\omega}[/math] Am I correct ? So, [math]x=\frac{\omega}{2\pi}\lambda t[/math] But, [math]\omega=\frac{\theta}{t}[/math] So, [math]x=\frac{\theta \lambda}{2\pi}[/math] [math]x=\frac{\frac{\theta}{t} \lambda t}{2\pi}[/math] without latex, you've got x = ((theta / t ) / 2pi) * lambda t omega doesnt drop the t down with 2pi to cancel you've got a fraction on a fraction t wouldnt cancel and everything should still be * lambda t Edited December 17, 2016 by DevilSolution Link to comment Share on other sites More sharing options...
Sriman Dutta Posted December 18, 2016 Author Share Posted December 18, 2016 So, [math]x=\frac{\omega}{2\pi}\lambda t[/math] But, [math]\omega=\frac{\theta}{t}[/math] So, [math]x=\frac{\theta \lambda}{2\pi}[/math] [math]x=\frac{\frac{[/size][/background]\theta}{t} \lambda t}{2\pi}[/math] without latex, you've got x = ((theta / t ) / 2pi) * lambda t omega doesnt drop the t down with 2pi to cancel you've got a fraction on a fraction t wouldnt cancel and everything should still be * lambda t Why?[math] x=vt=f\lambda t= \frac{\omega \lambda t}{2 \pi} = \frac{\theta \lambda t}{2 \pi t} = \frac{\theta \lambda}{2 \pi} [/math] Link to comment Share on other sites More sharing options...
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