MCQ
A wave travelling along uniform string represented by $Y=A \sin (\omega t-k x)$ is superimposed on another wave travelling along the same string represented by $Y=A \sin (\omega t+k x)$. The resultant is
  • A
    a wave travelling along $+x$ direction.
  • a standing wave having nodes at $x=\left(n-\frac{1}{2}\right) \frac{\lambda}{2}$, where $n=1,2,3, \ldots \ldots \ldots \ldots$
  • C
    a wave travelling along $-x$ direction.
  • D
    a standing wave having nodes at $x=\frac{n \lambda}{2}$, where $n=0,1,2, \ldots \ldots \ldots$.

Answer

Correct option: B.
a standing wave having nodes at $x=\left(n-\frac{1}{2}\right) \frac{\lambda}{2}$, where $n=1,2,3, \ldots \ldots \ldots \ldots$
(b) : According to principle of superposition, the resultant wave is
$
\begin{aligned}
& Y=A \sin (\omega t-k x)+A \sin (\omega t+k x) \\
& =2 A \sin \omega t \cos k x ......(i)
\end{aligned}
$
It is a standing wave equation.
For nodes, amplitude $Y=0$ i.e., $k x=0$
$
\Rightarrow \quad k x=\left(n-\frac{1}{2}\right) \pi ; n=1,2,3, \ldots
$
Since, $k=\frac{2 \pi}{\lambda}$; we get
$
x=\left(n-\frac{1}{2}\right) \frac{\lambda}{2} ; n=1,2,3 \ldots
$

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