Question

Two identical progressive waves travelling through a medium in opposite directions interfere with each other. Deduce the equation of the resultant wave. Hence show that the nodes and antinodes are equally spaced.

Answer 1

Hint: A stationary wave is produced only when a progressive wave and a resulting reflected wave is superimposed under boundary conditions. The principle of superposition of two waves is applied in order to find the total displacement of a particle due to the individual displacements of the two waves travelling in opposite directions. The concept of nodes and antinodes is used to prove that they are equally spaced by applying the conditions for the occurrence of a node and antinode.

Complete step by step answer:
The above problem revolves around the concept of stationary waves. First, let us look into the concept of stationary waves and how they are produced. Stationary waves are waves produced when two identical waves of same frequency and amplitude travelling with the same speed in opposite directions to each other superimpose each other and the resultant wave does not travel in either direction and is said to be stationary. This phenomenon is responsible for the formation of stationary waves or standing waves.

The waves are assumed to be produced on a string which is considered to be the medium of travel of the waves, where the creation of one wave results in the other, which means that in order to produce stationary waves there must be boundary effects applied on the string. This means that a finite string, that is, it has a finite medium in which it travels in.

This is mainly because the stationary waves cannot be produced by simply crossing two independent waves and hence the progressive wave and its subsequent reflected wave is considered. A progressing wave is one that travels from one point in the medium to another. Since, the stationary waves can only be produced in a finite medium, under boundary conditions this means that a boundary is considered.

This boundary is responsible for producing a reflected wave in accordance to Newton’s third law of motion which means that this resultant wave is formed when the independent wave produced hits the boundary wherein the boundary produces an equal and opposite reaction to the force applied on it leading to the formation of reflected wave.

Knowing this, let us now apply the concept of principle of superposition. Since the two waves are travelling in opposite directions to each other there will be a point in which both of these waves cross each other. This is called superimposing effect which leads to the formation of a resultant wave to be formed which is the standing wave. The equation for the standing wave is asked to be found out.

When a single particle in a string is considered and when there is a wave that is produced which is produced across this particle in the medium then there is some amount of displacement in the position of the particle. Since there are two waves considered here, there are two displacements respectively.

The equations for the displacements due to these waves are given by:
$y_1=A\sin (kx-\omega t)$ --------($1$)
$\Rightarrow y_2=A\sin (kx+\omega t)$ -------($2$)
We are considering that equation ($1$) is for the wave travelling in the positive direction on the x-axis while the equation ($2$) is for the wave travelling in the negative direction of the x-axis. It can be noted that the signs of equations ($1$) and ($2$) are reversed since they are travelling in opposite directions to each other.

Now, let us apply the principle of superposition to obtain the equation for the resultant wave. The superposition principle states that the resultant displacement is said to be equal to the algebraic sum of the individual displacements given to it by the two waves. Hence the resultant wave equation is obtained by simply adding equations ($1$) and ($2$). Here, $y$ is the total displacement of the waves. Hence we get:
$y=y_1+y_2$
$\Rightarrow y=A\sin (kx-\omega t)+A\sin (kx+\omega t)$
Since the waves are said to be identical and as per the conditions of stationary waves, we can say that the amplitude of the two waves are equal and hence this common tem is taken out.
$\Rightarrow y=A\left[\sin (kx-\omega t)+\sin (kx+\omega t)\right]$
We now apply the trigonometric formula for sine. It is given as:
$\sin A+\sin B=2\sin \left({\displaystyle \frac{a+b}{2}}\right)\cos \left({\displaystyle \frac{a-b}{2}}\right)$

By applying this above formula we get:
$\sin (kx-\omega t)+\sin (kx+\omega t)=2\sin \left({\displaystyle \frac{(kx-\omega t)+(kx+\omega t)}{2}}\right)\cos \left({\displaystyle \frac{(kx-\omega t)-(kx+\omega t)}{2}}\right)$
This is substituted in the equation for resultant wave to get:
$y=A\left[2\sin \left({\displaystyle \frac{(kx-\omega t)+(kx+\omega t)}{2}}\right)\cos \left({\displaystyle \frac{(kx-\omega t)-(kx+\omega t)}{2}}\right)\right]$
By simplifying the equation further we get:
$y=A\left[2\sin \left({\displaystyle \frac{kx-\omega t+kx+\omega t}{2}}\right)\cos \left({\displaystyle \frac{kx-\omega t-kx-\omega t}{2}}\right)\right]$
$\Rightarrow y=A\left[2\sin \left({\displaystyle \frac{kx+kx}{2}}\right)\cos \left({\displaystyle \frac{-\omega t-\omega t}{2}}\right)\right]$
$\Rightarrow y=A\left[2\sin \left({\displaystyle \frac{2kx}{2}}\right)\cos \left({\displaystyle \frac{-2\omega t}{2}}\right)\right]$
$\Rightarrow y=A\left[2\sin \left(kx\right)\cos \left(-\omega t\right)\right]$

Since, cosine function is an even function it obeys the rule that:
$\cos (-x)=\cos (x)$
Hence by applying this rule we get the equation to be:
$y=A\left[2\sin \left(kx\right)\cos \left(\omega t\right)\right]$ --------($3$)
Here, the variable $k$is known as the propagation constant and this is given by a formula
$k={\displaystyle \frac{2\pi }{\lambda }}$
This is substituted in the equation ($3$) to get:
$y=A\left[2\sin \left({\displaystyle \frac{2\pi x}{\lambda }}\right)\cos \left(\omega t\right)\right]$
$\Rightarrow y=2A\sin \left({\displaystyle \frac{2\pi x}{\lambda }}\right)\cos \left(\omega t\right)$ -------($4$)
This is the final equation for the resultant wave.

Now, we apply the concept of nodes and antinodes. When the resultant wave is produced it is said to be periodically produced which means that it repeats itself. There are certain particles that are present in the medium which do not undergo any displacement and are said to remain at rest. The positions where there is no displacement are called nodes. Correspondingly, there are other particles as well which are present in the medium which undergo any maximum displacement. The positions where this occurs are called antinodes.

We know that waves are periodic in nature and are expressed by either sine or cosine functions. The displacement for a simple wave is given by the equation:
$y=A\sin (\omega t)$ or $y=A\cos (\omega t)$
As per this principle, since the angular frequency is also said to be the same for the two waves, equation ($4$) becomes:
 $\Rightarrow y=\left(2A\sin \left({\displaystyle \frac{2\pi x}{\lambda }}\right)\right)\cos \left(\omega t\right)$
And hence the total amplitude of the resultant wave will be:
$\Rightarrow A^{\prime }=2A\sin \left({\displaystyle \frac{2\pi x}{\lambda }}\right)$

Since, the amplitude is said to be the maximum displacement of a particle in the medium from its mean position and since the node is said to have zero displacement we can say that amplitude is zero for the formation of nodes. Hence we have:
$2A\sin \left({\displaystyle \frac{2\pi x}{\lambda }}\right)=0$
$\Rightarrow \sin \left({\displaystyle \frac{2\pi x}{\lambda }}\right)=0$
$\Rightarrow \sin \left({\displaystyle \frac{2\pi x}{\lambda }}\right)=\sin (n\pi )$
$\Rightarrow {\displaystyle \frac{2\pi x}{\lambda }}=n\pi$
Cancelling out the common terms we get:
$\Rightarrow {\displaystyle \frac{2x}{\lambda }}=n$
The equation for $x$is hence:
$\Rightarrow x={\displaystyle \frac{n\lambda }{2}}$
Where, $n$ is a positive integer and hence $n=0,1,2,3...$
Hence, by replacing the value of $n$we can get the position of the nodes
$\Rightarrow x=0,{\displaystyle \frac{\lambda }{2}},\lambda ,{\displaystyle \frac{3\lambda }{2}}....$

Similarly we know that antinodes are formed when there is maximum displacement and hence the amplitude will be equivalent to $1$.
$\Rightarrow \sin \left({\displaystyle \frac{2\pi x}{\lambda }}\right)=\pm 1$
$\Rightarrow \sin \left({\displaystyle \frac{2\pi x}{\lambda }}\right)=\sin \left(\left(2n+1\right){\displaystyle \frac{\pi }{2}}\right)$
$\Rightarrow {\displaystyle \frac{2\pi x}{\lambda }}=\left(2n+1\right){\displaystyle \frac{\pi }{2}}$
Cancelling out the common terms we get:
$\Rightarrow {\displaystyle \frac{2x}{\lambda }}=\left(2n+1\right){\displaystyle \frac{1}{2}}$
By cross multiplying we get the equation for $x$:
$\Rightarrow x=\left(2n+1\right){\displaystyle \frac{\lambda }{4}}$
Since, $n$ is a positive integer and hence $n=0,1,2,3...$
We get:
$\therefore x={\displaystyle \frac{\lambda }{4}},{\displaystyle \frac{3\lambda }{4}},{\displaystyle \frac{5\lambda }{4}}...$
Hence we can see that the nodes and antinodes occur at periodic positions and hence the nodes and antinodes are placed at equal spaces because the difference between their positions is ${\displaystyle \frac{\pi }{4}}$ and is constant. This means that an alternating pattern of nodes and antinodes are formed at every ${\displaystyle \frac{\pi }{4}}$ interval.

Note: The shape of the waves crossing each other to not get affected and hence even after crossing each other they retain their shapes. However, at the exact position where they cross each other their shape may vary wherein their amplitude may increase or decrease. The superposition principle is applicable even for electromagnetic waves as well.