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 $t$.



1 Answer 1


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:

$x_0 = t (\lambda/ T)$


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

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

Then we get:

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


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