If we have a travelling wave of equation $y=5\sin(\omega t-\pi x +0.5)\:\rm m$. I know that the phase is 0.5 but how can I visualize this and what is the definition of phase difference between 2 points along the same wave that are $n$ meters apart in the same medium?
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You mean -πx/m in your sin argument. Draw two waves with their maxima 0.5m/π apart, at a moment of your choice; they travel at the same speed. – Cosmas Zachos Oct 30 '22 at 21:11
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@CosmasZachos What do the 2 waves you are talking about represent? – Youssef Mohamed Oct 30 '22 at 22:24
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They represent whatever you wish them to. – Cosmas Zachos Oct 31 '22 at 00:16
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The phase of an oscillation is the 'position' (usually expressed as an angle) that it has reached in its cycle. For example the phase of your wave at time $t$ and distance $x$ from some fixed point is $(\omega t -\pi[\text m^{-1}] x+0.5)$. It needs also to be stated whether the function is sine or cosine.
In your example, 0.5 is the phase constant.
The phase difference, $\Delta \phi$, at two points distance $\Delta x$ apart along the direction of propagation is the difference in these phase angles at any given time. Thus, $\Delta\phi=2\pi \frac {\Delta x}\lambda$. In your example, $\lambda = 2$m, so $\Delta \phi = 2\pi \frac {\Delta x}{2\text m}=\pi \ \text m^{-1} \Delta x$.
Philip Wood
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