Sriman Dutta Posted December 9, 2016 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
studiot Posted December 9, 2016 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
DevilSolution Posted December 17, 2016 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
Sriman Dutta Posted December 18, 2016 Author 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]
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