Question

A travelling wave represented by $y=A\sin (\omega t-kx)$ is superimposed on another wave represented by $y=A\sin (\omega t+kx)$. The resultant is

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

Answer 1

The correct option is A A standing wave having nodes at$x=(n+\frac{1}{2})\frac{\lambda }{2}$, where $n=0,1,2$

According to the principle of superposition, the resultant wave is
$y=asin(kx-\omega t)+asin(kx+\omega t)$
$=2asin\omega tcosx$ .....(i)

It represents a standing wave.
In the standing wave, there will be nodes (where amplitude is zero) and antinodes (where amplitude is largest).
From Eq. (i), the positions of nodes are given by
$sinkx=0⟹kx=n\pi ;n=0,1,2,....$
or $\frac{2\pi }{\lambda }x=n\pi ;0,1,2,....$
or $x=\frac{n\lambda }{2};n=0,1,2,...$

In the same way,
From Eq.(i), the positions of antinodes are given by$|sinkx|=1$
$⟹kx=(n+\frac{1}{2})\pi ;n=0,1,2,.....$
or $\frac{2\pi x}{\lambda }=(n+\frac{1}{2})\pi ;n=0,1,2,.....$
or $x=(n+\frac{1}{2})\frac{\lambda }{2};n=0,1,2,.....$