# Can the equation of total maximum amplitude $A_n=\sqrt{A_1^2+A_2^2+2A_1A_2\cos(\Delta\phi)}$ be used though the waves are not in the same line

Let $$S_1$$ and $$S_2$$ placed in the same point be the source of two waves which are propagating in the same line, also the phase differernce between the two waves $$\Delta\phi=0$$. Equation of the two waves is given by $$y_1=A_1\sin(\omega t-kx)$$ and $$y_2=A_2\sin(ωt-kx)$$ respectively.

Now at the distance $$x_1$$ from the sources, the equations of SHM of a particle become \begin{align} y_1=A_1\sin(\omega t-kx_1) \quad \text{(for wave 1)} \\ y_2=A_2\sin(\omega t-kx_1) \quad \text{(for wave 2)} \end{align} the resultant equation of SHM is given by just adding the two equation $$y_n=A_1\sin(\omega t-kx_1)+A_2\sin(\omega t-kx_2)$$

As written in my book the equation can repressented like $$y_n=A_n\sin(\omega t-kx_1-\theta)$$ where $$A_n$$ is the net maximum displacement due to the two waves and $$\theta$$ is phase difference. To find $$A_n$$ and $$\theta$$, we treat $$A_1$$ and $$A_2$$ as vector and consider the angle between them is same with the phase difference of two SHM $$\Delta\phi$$ .

According to the above $$A_n =\sqrt{A_1^2+A_2^2+2A_1A_2\cos(\Delta\phi)}.$$ The equation is good one when the waves are in the same line. Because if $$\Delta\phi=0$$ then the displacement for the each waves just add up and give the total displacement which also can be find by the above equation $$A_n= \sqrt{(A_1^2+A_2^2+2A_1A_2\cos(0)}= \sqrt{A_1^2+A_2^2+2A_1A_2} =A_1+A_2$$

Also the formula is effective when the waves are in the same line and phase of two SHM differ by $$\pi$$ as here the total displacement is the subtraction of two displacement due to the individual waves. I believe that the equation is valid for any other cases where the waves are in the same line though I do not find any reason why the angle between $$A_1$$ and $$A_2$$ displacement would be equal to the phase difference of two SHM due to the waves.

Though I have seen in the above two cases, it is undoutlessly applicable as when the phase difference is 0 the directions of the two displacement are same and when the phase difference is $$\pi$$, we can subtract the minor displacement from the mazor as the direction of the displacements are opposite . Those two cases show that we can take $$A_1$$ and $$A_2$$ as vector , also phase difference $$\Delta\phi$$ can be taken as angle between $$A_1$$ and $$A_2$$ . But when we think about any other cases where the phase difference is not $$0$$ or $$\pi$$ but the ange between the displacement is either $$0$$ or $$\pi$$ (when the particles go in the same direction the angle is $$0$$ and when they go opposite the angle is $$\pi$$) (note- I am assuming the waves are in the same line)

Then in those cases , why do we use the phase difference $$\Delta\phi$$ as the angle between $$A_1$$ and $$A_2$$ insteat of $$0$$ and $$\pi$$.

Another problem with the equation i find when i think such a case where the waves are not in the same line.

Let the equation of two waves be $$y_1=A_1\sin(\omega t-kx)$$ and $$y_2=A-2\sin(\omega t-kx)$$ respectively. Now the two waves superpose at point $$P$$ with the angle $$\pi/2$$ means the waves are perpendicular with each other. let the distance travelled by the first wave to reach point $$P$$ be equal to distance travelled by the second wave to reach point $$P$$. If the distance is $$x_1$$
then the equation of of SHM of a particle on point $$P$$ (the point of superposition) become \begin{align} y_1=A_1\sin(\omega t-kx_1) \quad \text{(for wave 1)} \\ y_2=A_2\sin(\omega t-kx_1) \quad \text{(for wave 2)} \end{align}

We can see clearly that the phase difference between two SHM $$\Delta\phi$$ is $$0$$ as the path difference $$∆x$$ is $$0$$.

So according to my book the equation of resultant SHM is given by
\begin{align} y_n&=y_1+y_2 \\ &=A_1\sin(\omega t-kx_1)+A_2\sin(\omega t-kx_2) \\ &=A_n\sin(\omega t-kx_1-θ) \end{align}

And
\begin{align} A_n&=\sqrt{(A_1^2+A_2^2+2A_1A_2\cos(\Delta\phi)} \\ &=\sqrt{A_1^2+A_2^2+2A_1A_2\cos(0)} \quad \text{(as phase difference is 0)} \\ &=\sqrt{A_1^2+A_2^2+2A_1A_2)} \\ &=A_1+A_2 \end{align}

But if we imagine the the situation we will find the angle between $$A_1$$ and $$A_2$$ is $$\pi/2$$ as the waves superpose by the angle of $$\pi/2$$. So the value of $$A_n$$ should be equal to \begin{align} A_n&= \sqrt{A_1^2+A_2^2+2A_1A_2\cos(π/2)} \\ &=\sqrt{A_1^2+A_2^2} \quad \text{(as two SHM are same phase so when y_1=A_1, y_2=A_2)} \end{align}

That does not match with the above which i got using my book formula.

• Can the equation of total maximum amplitude $$Aₙ=\sqrt{A_1^2+A_2^2+2A_1A_2\cos(\Delta\phi)}$$ be used though the waves are not in the same line.
• So let say you have two waves $A_{1}e^{j\mathbf{k1}.\mathbf{r}-\omega t}$, $A_{2}e^{j\mathbf{k2}.\mathbf{r}-\omega t}$, with $\vert \mathbf{k_{1}} \vert=\vert \mathbf{ k_{2} }\vert$=k, the angle $(\mathbf{k_{1}},\mathbf{k_{2}})= \theta$, what exactly do you want to know? – user8736288 Apr 6 '20 at 15:17
For your first question I assume what you want is a derivation of your formula. What you want is to express the real part of $$A_{1}e^{j(-kx+\omega t)}+A_{2}e^{j(-kx+\omega t+\phi)}$$ as new expression which is the real part of $$A_{n}e^{-j(kx+\omega t+ \psi)}$$ so you may simply write: $$A_{1}e^{j(kx-\omega t)}+A_{2}e^{j(kx-\omega t+\phi)} = (A_{1} +A_{2}e^{j\phi})e^{j(kx-\omega t)}$$ Now since the complex $$(A_{1} +A_{2}e^{j\phi})=(A_{1}+A_{2}\cos(\phi)+ j A_{2}sin(\phi))$$ its modulus is indeed $$A_{n}=\sqrt{A_{1}^{2} +A_{2}^{2}+ 2A_{1}A_{2}\cos(\phi)}$$ and its argument is $$\psi=\left( \frac {A_{2}\sin(\phi)}{A_{1}+A_{2}\cos(\phi)} \right)$$.
For your second question, beware that the angle difference in the phasors representing your waves cannot be identified with the angle between the vectors $$\mathbf{k_{1}}$$ and $$\mathbf{k_{2}}$$ (I assume $$\vert \mathbf{k_{1}} \vert = \vert \mathbf{k_{2}} \vert =k )$$ that is, the angle between the vectors defining the direction of propagation of 2 waves $$A_{1}e^{j(\mathbf{k_{1}}\mathbf{r} -\omega t)}$$ and $$A_{1}e^{j(\mathbf{k_{2}}\mathbf{r} -\omega t)}$$. Assuming you look at the field in a particular direction such that $$\mathbf{r}=r\mathbf{i}$$, and denoting $$\theta_{1}$$ the angle between $$\mathbf{i}$$ and $$\mathbf{k_{1}}$$, $$\theta_{2}$$ the angle between $$\mathbf{i}$$ and $$\mathbf{k_{2}}$$, we have: $$\mathbf{k_{1}}.\mathbf{r}= k\cos(\theta_{1})r=k_{1}r$$ $$\mathbf{k_{2}}.\mathbf{r}= k\cos(\theta_{2})r=k_{2}r$$ The resulting field along $$\mathbf{i}$$ is: $$A_{1}e^{j(k_{1}r -\omega t)} + A_{2}e^{j(k_{2}r -\omega t)}$$ So the difference of phase between the two phasors is not constant, it is equal to $$(k_{2}-k_{1})r$$ and depends on $$r$$, your formula may thus not be so useful in this case. Of course in the particular case when $$\theta_{1} = \theta_{2}=\theta$$ then $$k_1= k_2= k\cos(\theta)$$ and the resulting fied simply writes: $$(A_{1}+A_{2})e^{j(kcos(\theta)r -\omega t)}$$ Hope it helps.