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?

Question

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.

Please explain the two things

• Why do we take the angle between vector A₁ and A₂ same as phase difference of two SHM of medium partical at the point of superposition.
• 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.

Answer 1

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.