# Trying to prove the wave equation from circle

Imagine one of the point in the wave. It is in oscillation. So its displacement can be written as $$Y = A\sin(\theta)$$ where $$\theta= \omega t$$. $$Y(t)=A\sin(\omega t) \tag{1}$$

Time for one wave length $$\lambda$$ to travel is $$T$$. So for $$x$$ position, $$Tx/\lambda$$ will be the time. Substituting it in (1), we get $$Y(x)=A\sin(2\pi x/\lambda)$$ $$Y(x)=A\sin(kx) \tag{2}$$

Can we find $$Y(x,t)=A\sin(kx+\phi-\omega t)$$ from a circle with radius as the amplitude $$A$$?

What I find difficult in doing it is the following at $$t =0$$ particle makes an angle $$\phi$$ (phase constant). Equation (2) is the displacement at time $$t$$. • You should take some time to learn to use MathJax notation. It makes equations more readable. math.meta.stackexchange.com/questions/5020/… Jul 23 at 12:05
• That's so cool.Will learn it. Jul 23 at 12:27

In the first equation of $$Y(t)$$ you are talking about the equation of $$y$$ at one determinate point of the wave, for example $$x=0$$ or $$x=2$$. The problem is that you can not substitute $$t$$ by $$xT/\lambda$$. Because this isn't true. When talking about the wave you can not relate $$x$$ and t as to know $$Y$$ you have to know both $$x$$ and $$t$$.

I will explain the deduction so you can understand. Imagine the wave when $$t=0$$, so the formula Will be:

$$Y(x,0) = A \sin{(kx)}$$

But as time passes the wave moves. What changes is the initial position $$x_0$$ of the wave. So at time t:

$$Y(x,t) = A \sin{(k(x-x_0))}$$

What we know is that when $$t=T$$, the $$x_0$$ becomes $$\lambda$$, so:

$$x_0 = t (\lambda/ T)$$

So,

\begin{align}Y(x,t) &= A \sin{(k(x-x_0))}\\ & = A \sin{(k(x - t (\lambda/T)))}\\ &= A \sin{(kx-t.k.\lambda/T)}\end{align}

And, $$k.\lambda/T = (2.\pi.\lambda)/(\lambda.T) = 2.\pi/T = \omega$$

Then we get:

$$Y(x,t) = A \sin{( kx-\omega t)}$$