# Interference of two wave

We can produce a standing wave by superposition of two waves:one incident $$y_1=A \sin{(kx-wt)}$$and the other one is reflected $$y_2=A \sin{(kx+wt)}$$ ;

(1)According to my teacher, if the two waves has phase difference of $$180°$$ ,then the resulting wave will be $$y_1+y_2$$;

(2)From textbook i know that the resulting wave from $$y_1=A \sin{(kx-wt)}$$and $$y_2=A\sin{(kx-wt +\phi)}$$will be $$y=2A{\cos{(\frac{\phi}{2})}}{\sin{(kx-wt+\frac{\phi}{2})}}$$

• You want to relate the two? Put $\phi=180°$ – Harshit Joshi Jan 7 '19 at 18:56

In the first case, with $$y_1=A\sin(kx-wt)$$ and $$y_2=A\sin(kx+wt)$$ you get a standing wave, as the sum is $$y_1+y_2=2A\sin(kx)\cos(wt)$$. This is becasue the two waves are propagating in opposite direction. In the second case, you don't have standing waves becasue the waves are propagating in the same direction. Anyway, it's just simple trigonometry. From the relations $$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha$$ $$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\sin\beta\cos\alpha$$ you get $$\sin(\alpha+\beta)+\sin(\alpha-\beta)=2\sin\alpha\cos\beta$$ Now, let $$\gamma=\alpha+\beta$$ and $$\delta=\alpha-\beta$$, the you have $$\alpha=\frac{\gamma+\delta}{2}$$ $$\beta=\frac{\gamma-\delta}{2}$$ so $$\sin\gamma+\sin\delta=2\sin\frac{\gamma+\delta}{2}\cos\frac{\gamma-\delta}{2}$$

Then in case (1) you have $$\gamma=kx-wt$$ and $$\delta=kx+wt$$ and from the formula above you get $$y_1+y_2=2A\sin(kx)\cos(wt)$$ In case (2) you have $$\gamma=kx-wt$$ and $$\delta=kw-wt+\phi$$. Putting this in the abowe equation you get $$\sin(kx-wt)+\sin(kw-wt+\phi)=2\sin\left(kw-wt+\frac{\phi}{2}\right)\cos\left(\frac{\phi}{2}\right)$$ I have considered $$A=1$$.

The use of phasors can help in such a case?

In the diagram the reference phasor $$WX$$ represents $$A\sin(kx-\omega t)$$ with amplitude $$A$$ at some position $$x$$ and phasor $$WY$$ represents $$A\sin(kx-\omega t+\phi)$$ at the same position with the same amplitude $$A$$ and leading $$A\sin(kx-\omega t)$$ by $$\phi$$.

The diagram shows immediately that the resultant phasor $$WY$$ leads the reference phasor by $$\frac \phi 2$$ and so is of the form $$\sin(kx-\omega t+\frac \phi 2)$$

The amplitude of the resultant phasor is the length of $$WY$$ which can be found by application of the cosine rule on triangle $$WXY$$

$$WY^2 = A^2 + A^2 - 2 \,A\,A\cos (\pi - \phi) = A^2 \,2(1-\cos^2 \phi)= 4A^2\cos^2 \left (\frac \phi 2 \right)$$

$$\Rightarrow WY = 2A\cos \left (\frac \phi 2 \right)$$

The resultant phasor has an amplitude of $$2A\cos \left (\frac \phi 2 \right)$$ so can be written as $$2A\cos \left (\frac \phi 2 \right)\sin(kx-\omega t+\frac \phi 2)$$

Response to a question asked by the OP.

The standing wave can be dealt with in this way.

The $$A\sin(kx +\omega t)$$ variation can be thought of as a $$A\sin(\omega t + \phi)$$ variation at a position $$x$$ where $$\phi = kx$$.
This can be drawn as a phasor which is $$\phi$$ in advance of the reference phasor $$A\sin{\omega t}$$ phasor.

The $$A\sin(kx - \omega t)$$ variation at the same position $$x$$ is a little trickier to deal with in the following way.

$$A\sin(kx - \omega t) = - A \sin(\omega t - kx) = A \sin(\omega t - kx - \pi) = A \sin(\omega t - (\phi + \pi))$$

This is a phasor which lags behind the reference phasor $$A\sin \omega t$$ by $$\phi + \pi$$

So as you move position $$x$$ the phase $$\phi$$ changes to give a different but constant amplitude for the resultant phasor.

When $$x=0 \Rightarrow \phi =0$$ the resultant is $$0$$ and when $$x= \frac{\pi}{2k} \left (= \frac{\lambda}{4}\right )$$ the resultant is $$2A$$.

The resultant phasor is $$\frac \pi 2$$ ahead of the reference $$\sin\omega t$$ phasor ie it is $$\cos \omega t$$ and has an amplitude of $$2A\sin \phi \Rightarrow 2A\sin\phi\cos \omega t$$

• Can we use this method for the first case?and i want to be confirmed about $~$ $(kx+wt)$,why not it is $~$$(kx-wt)$ – Raihan Amin Jan 8 '19 at 12:19