Question

Explain the formation of stationary waves by analytical method. Show that nodes and antinodes are equally spaced in stationary waves.

Answer 1

Hint: To solve this problem, we must remember the basic formula of a simple harmonic wave, which is, $X=A\sin [\omega t\pm \phi ]$ . The condition for a stationary wave, which is, that the wave equation must be independent of the position term. We must also remember that at an antinode, is where the amplitude is maximum and a node is where the amplitude is minimum.

Step by step solution:
The analytical treatment means to analytically solve the whole equation using the conditions. Hence, for the analytical treatment of stationary waves, let’s start by considering two simple harmonic waves progressing towards each other, coming from opposite directions. These two stationary waves have the same amplitude (a), same frequency $(\upsilon )$ and same time period (T).
Let the first wave directed along the positive x axis be: ${{X}_{1}}=a\sin [2\pi (\upsilon t+\dfrac{x}{\lambda })]$ and the wave directed along the negative x axis be: ${{X}_{2}}=a\sin [2\pi (\upsilon t-\dfrac{x}{\lambda })].$

Using the principle of superposition, the resultant displacement of the stationary wave will be (X). That is, $X={{X}_{1}}+{{X}_{2}}.$

Therefore, resulting wave becomes,$X=a\sin [2\pi (\upsilon t+\dfrac{x}{\lambda })]+\alpha \sin [2\pi (\upsilon t-\dfrac{x}{\lambda })]$

Using the $\sin A+\sin B$ relation, where $\sin A+\sin B$ becomes; $\sin A+\sin B=2\sin (\dfrac{A+B}{2})\cos (\dfrac{A-B}{2})$.

Hence, using the above relation, the value of superposition becomes; $X=2a\sin [\dfrac{2\pi }{2}(\upsilon t+\dfrac{x}{\lambda }+\upsilon t-\dfrac{x}{\lambda })].\cos [\dfrac{2\pi }{2}(\upsilon t+\dfrac{x}{\lambda }-\upsilon t+\dfrac{x}{\lambda })]$
This implies, $X=2a\sin [2\pi \upsilon t].\cos [2\pi \dfrac{x}{\lambda }]$.

Therefore, $X=A\sin [2\pi \upsilon t]$, where the value of A is, $A=2a\cos [\dfrac{2\pi x}{\lambda }]$. This value of A, is the amplitude of the resulting superimposed wave.
Hence, $X=A\sin [\omega t]$, where$(\omega )$ is the angular frequency of the superimposed wave. Thus, the expression becomes a simple harmonic wave having the same time period. However, the amplitude changes as compared to the initial two waves. This resultant superimposed wave is a stationary wave.

To find the conditions for antinodes, let’s first understand what an antinode is. The points in the medium that vibrate with maximum amplitude where the wave is propagating are known as antinodes. Hence, the basic condition would be, points where A is maximum, for the resultant superimposed wave.

Hence, $\left| A \right|=2a\Rightarrow A=\pm 2a.$ That is, $2a\cos [\dfrac{2\pi x}{\lambda }]=\pm 2a\Rightarrow \cos [\dfrac{2\pi x}{\lambda }]=\pm 1.$

We know that, $\cos n\pi =\pm 1,$ where n can be any integer. Therefore, the above equation becomes, $\cos [\dfrac{2\pi x}{\lambda }]=\cos n\pi \Rightarrow \dfrac{2\pi x}{\lambda }=n\pi \Rightarrow x=\dfrac{n\lambda }{2}.$

Hence, putting in n=0, we get, x=0 and upon putting n=1, we get, $x=\dfrac{\lambda }{2}$ and for n=-1, we get, $x=-\dfrac{\lambda }{2}$ and so on. Hence, these values of x are the points of antinodes in the resultant superimposed wave.

Similarly, to find the conditions for nodes, let’s understand what is a node. The points in the medium that vibrate with minimum amplitude where the wave is propagating are known as nodes. Hence, the basic condition would be, points where A is minimum, for the resultant superimposed wave.

Hence, $A=0.$ That is, $2a\cos [\dfrac{2\pi x}{\lambda }]=0\Rightarrow \cos [\dfrac{2\pi x}{\lambda }]=0.$

We know that, $\cos \dfrac{(2n\pm 1)\pi }{2}=0,$where n can be any integer. Therefore, the above equation becomes, $\cos [\dfrac{2\pi x}{\lambda }]=\cos \dfrac{(2n\pm 1)\pi }{2}\Rightarrow \dfrac{2\pi x}{\lambda }=\dfrac{(2n\pm 1)\pi }{2}\Rightarrow x=\dfrac{(2n\pm 1)\lambda }{4}.$

Hence, putting in n=0, we get, $x=\pm \dfrac{\lambda }{4}$ and upon putting n=1, we get, $x=\dfrac{3\lambda }{4}$ and $x=\dfrac{\lambda }{4}$, and so on. Hence, these values of x are the points of nodes in the resultant superimposed wave.

Note:
The absence of the x term in the amplitude of the superimposed wave implies that the amplitude of the resultant superimposed wave does not move forward or backward, due to +x or -x respectively. Hence, the wave is a stationary wave.

The points, x=0, $\pm \dfrac{\lambda }{2},\pm \lambda ,\pm \dfrac{3\lambda }{2}...$ are the antinodes, while the points, \[x=\pm \dfrac{\lambda }{4},\pm \dfrac{3\lambda }{4},\pm \dfrac{5\lambda }{4}...\] are the nodes. The positive and negative sign here merely denote the position of the point with respect to the center in a number line.

Another important point to remember is that, the difference between any two successive nodes, or two successive antinodes is always equal to \[\dfrac{\lambda }{2}.\]