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, 20169 yr by Sriman Dutta

What happened to k in your working?

 

It is in there because ther are multiple solutions to the equation v = x/t

So, will it be: v=kx/t ?

Can anyone guide me?

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, 20169 yr by DevilSolution

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]