Question

A travelling wave represented by $y=A\sin (\Omega \text{t-Kx)}$ is superimposed on another wave represented by $y=A\sin (\Omega \text{t+Kx)}$. The resultant is

• A. A standing wave having nodes at
  $x=(n+\frac{1}{2})\frac{\lambda }{2},n=0,1,2$
• B. A standing wave having nodes at $\frac{n\lambda }{2};n=0,1,2$
• C. A wave travelling along $+x$ direction
  
• D. A wave travelling along $-\text{x}$ direction
  

Answer 1

The correct option is A A standing wave having nodes at

$x=(n+\frac{1}{2})\frac{\lambda }{2},n=0,1,2$
Given:
$y_1=\text{A}\sin (\Omega \text{t-kx)}$
$y_2=\text{A}\sin (\Omega \text{t+kx)}$
We know that,
Resultant wave
$y=y_1+y_2=\text{A}\sin (\Omega \text{t-kx)}+\text{A}\sin (\Omega \text{t+kx)}$
$y=2\text{A}\sin \Omega t\cos \text{kx}$
Clearly it is equation of standing wave.
For the position of nodes $y=0$
i.e. $x=(2n+1)\frac{\lambda }{4}$
$\Rightarrow x=(n+\frac{1}{2})\frac{\lambda }{2}\text{,}$
Where $n=0,1,2,3$
Hence , option $\left(A\right)$ is correct.