Question

Two travelling waves, ${{y}_{1}}=A\sin \left[ k\left( x+ct \right) \right]$ and ${{y}_{2}}=A\sin \left[ k\left( x-ct \right) \right]$ are superposed on string. The distance between adjacent antinode is:
$\begin{align}
  & A)\dfrac{ct}{\pi } \\
 & B)\dfrac{ct}{2\pi } \\
 & C)\dfrac{\pi }{2k} \\
 & D)\dfrac{k}{\pi } \\
 & E)\dfrac{\pi }{k} \\
\end{align}$

Answer 1

Hint: To solve this question, we need to find the resultant displacement of the two waves given in the question after the superposition of the waves. Then we will find the position of the two adjacent antinodes and then their difference to find the distance between them.

Complete step by step answer:
The superposition of waves can be defined as the resultant displacement of more than one waves in a medium at one particular point is the vector sum of the displacements of all the waves.
In the question two travelling waves are given as,
${{y}_{1}}=A\sin \left[ k\left( x+ct \right) \right]$ and ${{y}_{2}}=A\sin \left[ k\left( x-ct \right) \right]$
The resultant of the displacement of the two waves after superposition can be defined as,
$\begin{align}
  & y={{y}_{1}}+{{y}_{2}} \\
 & y=A\sin \left[ k\left( x+ct \right) \right]+~A\sin \left[ k\left( x-ct \right) \right] \\
 & y=A\left[ \sin \left\{ k\left( x+ct \right) \right\}+\sin \left\{ k\left( x-ct \right) \right\} \right] \\
\end{align}$
Using the laws of trigonometry, we can simplify the above equation as,
$\begin{align}
  & y=2A\sin \dfrac{kx+kct+kx-kct}{2}\cos \dfrac{kx+kct-kx+kct}{2} \\
 & y=2A\sin kx\cos kct \\
\end{align}$
A standing wave node is a point where the amplitude of the wave is zero and antinode is a point where the amplitude of the wave is maximum.
Now, let the first antinode be at position ${{x}_{1}}$ .
So, we can write that,
$\begin{align}
  & \sin k{{x}_{1}}=1 \\
 & \Rightarrow \sin k{{x}_{1}}=\sin \dfrac{\pi }{2} \\
 & \Rightarrow k{{x}_{1}}=\dfrac{\pi }{2} \\
 & \therefore {{x}_{1}}=\dfrac{\pi }{2k} \\
\end{align}$
Now, let the second antinode is found at ${{x}_{2}}$
So, we can write,
$\begin{align}
  & \sin k{{x}_{2}}=-1 \\
 & \Rightarrow \sin k{{x}_{2}}=\sin \dfrac{3\pi }{2} \\
 & \Rightarrow k{{x}_{2}}=\dfrac{3\pi }{2} \\
 & \therefore {{x}_{2}}=\dfrac{3\pi }{2k} \\
\end{align}$
So, the distance between two adjacent antinodes can be given as,
$\begin{align}
  & \Delta x={{x}_{2}}-{{x}_{1}} \\
 & \Rightarrow \Delta x=\dfrac{3\pi }{2k}-\dfrac{\pi }{2k} \\
 & \therefore \Delta x=\dfrac{\pi }{k} \\
\end{align}$
So, the distance between two adjacent antinode, when we superpose ${{y}_{1}}=A\sin \left[ k\left( x+ct \right) \right]$ and ${{y}_{2}}=A\sin \left[ k\left( x-ct \right) \right]$ is $\dfrac{\pi }{k}$ .

So, the correct answer is “Option E”.

Note: To find the position of the antinode we have put the value of the $\sin kx$ as 1 and -1.
To find the position of the nodes we need to put the value of the $\sin kx$ as zero. So, the value of the position will be either zero or integral multiple of $\dfrac{\pi }{k}$ where the distance between two nodes will be the same as the distance between two antinodes.