Solved Problem on Waves

Two progressive waves of the same range and same frequency have a phase difference equal to φ. Determine the expression of the wave resulting from the superposition of these two waves.

Problem data:
  • Waves amplitude:    y1 = y2 = a;
  • Angular frequency of waves:    ω1 = ω2 = ω;
  • Phase difference between the two waves:    φ.
Problem diagram:

We consider two sine waves, one (in red) has an initial phase equal to zero, and the other (in blue) has a phase difference equal to φ relative to the first wave, and both with amplitude a (Figure 1)

Figure 1


The wave function is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {y(x,t)=A\cos (kx-\omega t)} \end{gather} \]
Writing the wave functions for both waves
\[ \begin{gather} y_{1}(x,t)=a\cos (kx-\omega t) \\[5pt] \text{and}\\[5pt] y_{2}(x,t)=a\cos (kx-\omega t+\varphi ) \end{gather} \]
adding the two waves
\[ \begin{gather} y(x,t)=a\cos (kx-\omega t)+a\cos (kx-\omega t+\varphi)\\[5pt] y(x,t)=a[\cos (kx-\omega t)+\cos (kx-\omega t+\varphi)] \tag{I} \end{gather} \]
Note: From one of the Prostapheresis Formulas
\[ \begin{gather} \cos a+\cos b=2\operatorname{sen}\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right) \end{gather} \]

Doing the following associations   \( a=kx-\omega t \)   and   \( b=kx-\omega t+\varphi \),   so the expression (I) can be rewritten
\[ \begin{gather} y(x,t)=a\left[2\operatorname{sen}\left(\frac{kx-\omega t+kx-\omega t+\varphi }{2}\right)\cos \left(\frac{kx-\omega t-(kx-\omega t+\varphi)}{2}\right)\right]\\[5pt] y(x,t)=2a\operatorname{sen}\left(\frac{2kx-2\omega t+\varphi }{2}\right)\cos \left(\frac{kx-\omega t-kx+\omega t-\varphi}{2}\right)\\[5pt] y(x,t)=2a\operatorname{sen}\left(kx-\omega t+\frac{\varphi}{2}\right)\cos \left(-{\frac{\varphi }{2}}\right) \end{gather} \]
the cosine is an even function,   \( f(x)=f(-x) \),   we have   \( \cos \left(-{\frac{\varphi }{2}}\right)=\cos \left(\frac{\varphi}{2}\right) \)
\[ \begin{gather} y(x,t)=2a\cos \left(\frac{\varphi}{2}\right)\operatorname{sen}\left(kx-\omega t+\frac{\varphi}{2}\right) \end{gather} \]
setting   \( A\equiv 2a\cos \left(\frac{\varphi }{2}\right) \)
\[ \begin{gather} \bbox[#FFCCCC,10px] {y(x,t)=A\operatorname{sen}\left(kx-\omega t+\frac{\varphi }{2}\right)} \end{gather} \]

Note: The amplitude A of the resultant wave from the superposition will depend on the phase difference φ.
\[ \begin{gather} A=2a\cos \left(\frac{\varphi }{2}\right) \end{gather} \]
the term 2a is constant, and the absolute value of cosine ranges from 0 to 1.
Thus the amplitude will be maximum for
\[ \begin{gather} A=2a\underbrace{\cos \left(\frac{\varphi }{2}\right)}_{1}\Rightarrow A=2a \end{gather} \]
this happens when the angle of phase difference is equal to
\[ \begin{gather} 2a\cos \left(\frac{\varphi }{2}\right)=2a\\[5pt] \cos\left(\frac{\varphi }{2}\right)=\frac{2a}{2a}\\[5pt] \cos \left(\frac{\varphi}{2}\right)=1\\[5pt] \frac{\varphi }{2}=\arccos 1 \end{gather} \]
the angles for which the cosine is equal to one are those equal to 0, 2π, 4π, 6π,..., 2nπ
\[ \qquad\qquad\qquad\qquad\quad \begin{array}{c} \dfrac{\varphi }{2}=2n\pi & \phantom{} & \phantom{} \\ \varphi =2.2n\pi & \phantom{} & \phantom{} \\ \varphi =4n\pi & \text{,} & n=0,1,2,3,... \end{array} \]
In this case, the two waves have no phase difference (are in phase φ = 0, and both waves coincide and produce a resultant wave from the superposition with maximum amplitude, twice the original waves (Figure 2).

Figure 2

From this point, while the phase difference between the waves increases, the wave amplitude formed by the superposition decreases, in Figure 3 we see two examples when the lag is valid   \( \varphi =\frac{2\pi }{5} \)   and   \( \varphi =\frac{3\pi }{5} \).

Figure 3

The amplitude of the wave formed from the superposition decreases until reaches a certain phase difference it will be equal to the amplitude of the two superposition waves, this happens when
\[ \begin{gather} 2a\cos \left(\frac{\varphi }{2}\right)=a\\[5pt] \cos\left(\frac{\varphi }{2}\right)=\frac{\cancel{a}}{2\cancel{a}}\\[5pt] \cos \left(\frac{\varphi}{2}\right)=\frac{1}{2}\\[5pt] \frac{\varphi }{2}=\arccos\frac{1}{2} \end{gather} \]
The angles for which the cosine is equal to   \( \frac{1}{2} \),   considering only the first quadrant of the unit circle, is equal to   \( \frac{\pi }{3} \)
\[ \begin{gather} \frac{\varphi }{2}=\frac{\pi }{3}\\[5pt] \varphi =\frac{2\pi}{3} \end{gather} \]
In Figure 4, we see that the resultant wave from the superposition (in black) is the same amplitude as the superposed waves.

Figure 4

Continuing the increase in the phase difference between the waves, the result of the superposition has an amplitude less than the amplitude of the superposed waves, Figure 5 for a phase difference of   \( \varphi =\frac{4\pi }{5} \).

Figure 5

Further increasing the phase difference, the amplitude should continue to decrease until it reaches a value equal to zero, A = 0, this happens when the phase difference is equal to
\[ \begin{gather} 2a\cos \left(\frac{\varphi }{2}\right)=0\\[5pt] \cos\left(\frac{\varphi }{2}\right)=\frac{0}{2a}\\[5pt] \cos \left(\frac{\varphi}{2}\right)=0\\[5pt] \frac{\varphi }{2}=\arccos 0 \end{gather} \]
the angles for which the cosine is equal to zero are those equal to  \( \frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},...,\frac{(2n+1)\pi}{2},... \)
\[ \qquad\qquad\qquad\qquad\quad \begin{array}{c} \dfrac{\varphi}{2}=\dfrac{(2n+1)\pi }{2} & \phantom{} & \phantom{} \\ \varphi=\cancel{2}.\dfrac{(2n+1)\pi}{\cancel{2}} & \phantom{} & \phantom{} \\ \varphi=(2n+1)\pi & \text{,} & n=0,1,2,3,... \end{array} \]
In Figure 6, we have two superposed waves of   \( \varphi =\pi \),   the two waves produce a resultant wave that does not oscillate, having amplitude zero to their entire length.

Figure 6