# Explanation as to why the sum of two sinusoidal waves, differing by only phase, can be represented by $2y_{m}\cos(\frac{1}{2} \Phi)$

How does the addition of two waves, differing only by phase, collapse to $2y_{m}\cos(\frac{1}{2} \Phi)$?

Wouldn't the $\omega$ component of the wave still come into play given that it determines the period of the wave? i.e. $\omega=2\pi f$ and hence, $T = \frac{2\pi}{\omega}$.

I don't know where your formula comes from and nor what the symbols exactly mean but if $\Phi$ is the phase difference, then it is obviously wrong in the case where $\Phi = 0$ where one expects simply a propagating wave with twice the amplitude.

To get the general result, let us consider two waves $y_{1,2} = y_m \cos(\omega t - kx + \varphi_{1,2})$. Now let us introduce $\Phi = \varphi_1-\varphi_2$ and $\phi = \varphi_1+\varphi_2$. We get that $\varphi_1 = (\Phi+\phi)/2$ and $\varphi_2 = (\phi-\Phi)/2$. This enables to rewrite $y_{1,2} = y_m \cos(\omega t - kx + \phi/2 \pm \Phi/2)$.

Next, let us consider the following trigonometric identities $\cos(a+b) = \cos a \cos b - \sin a \sin b$ and $\cos(a-b) = \cos a \cos b + \sin a \sin b$. We get from them that $\cos(a+b)+\cos(a-b)=2\cos a \cos b$. Using $a = \omega t-kx + \phi/2$ and $b = \Phi/2$, we get for the two waves $y_1+y_2 = 2y_m \cos(\omega t-kx + \phi/2)\cos(\Phi/2)$. This equation has the nice property of giving back the right wave with a double amplitude in case of no phase difference.

A relatively simple way is to draw a phasor diagram.

Then use the cosine rule.

$y^2 = y_m^2 + y_m^2 + 2 y^2_m\cos \Phi = 2y^2_m(1+\cos \Phi) = 4y^2_m \cos^2 \frac \Phi 2 \Rightarrow y = 2y_m \cos \frac \Phi 2$

• Phasors are a tool that I didn't really 'get' in school and steadfastly ignored for as long as I could, but they've grown on me in recent years. Once you're comfortable with them you start seeing them in the most surprising places. – dmckee --- ex-moderator kitten May 8 '16 at 17:06

Let me tell about the case of Double-Slit experiment:

Let the wavefunction as

$$\psi(x,t)= A_0\cos\omega\left( t-\frac xv\right) \;.$$

At a point $P\,,$ at distances $x_1$ and $x_2$ from the first and two slits, the wavefunction can be expressed as:

\begin{align}\psi_p(t) &=A_0\cos\omega\left( t-\frac {x_1}v\right)+ A_0\cos\omega\left( t-\frac {x_2}v\right)\\ &= 2A_0\cos\omega t\cos\left[\frac\omega{2v}(x_2-x_1)\right]\end{align}\;.\tag 1

Using $\frac{2\pi v}\omega= \lambda$ in $(1)\,$ we get

$$\psi_p(t)= 2A_0\cos\omega t\cos\left[\frac{\pi}{\lambda}(x_2-x_1)\right]\;.\tag 2$$

Now, phase difference $$\Phi= \frac{2\pi}{\lambda}\,(x_2-x_1)\;;$$

using this in $(2)\,,$ we get

\begin{align}\psi_p(t)&= 2A_0\cos\omega t\cos\left(\frac{\Phi}2\right)\\ &= 2y_m\cos\left(\frac\Phi 2\right)\,,\end{align}

where

$y_m= A_0\cos\omega t\;.$

• Okay, that's my interpretation of $y_m$ although OP didn't clear the terminologies. – user36790 May 8 '16 at 11:06