Solved Problem on Waves
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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.
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)
Solution
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.
Thus the amplitude will be maximum for
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
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
Figure 6
\[
\begin{gather}
A=2a\cos \left(\frac{\varphi }{2}\right)
\end{gather}
\]
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).
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 \( \varphi =\frac{3\pi }{5} \).
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.
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 \( \varphi =\frac{4\pi }{5} \).
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.
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .