MHT CET · Physics · Waves and Sound
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.
- B A standing wave having nodes at \(x=\left(n+\frac{1}{2}\right) \frac{\lambda}{2}\), where \(n=0,1,2,3 \ldots \ldots\)
- C A wave travelling along \(-x\) direction
- D A standing wave having nodes at \(x=\frac{n \lambda}{2}\), where \(n=0,1,2,3 \ldots \ldots\).
Answer & Solution
Correct Answer
(B) A standing wave having nodes at \(x=\left(n+\frac{1}{2}\right) \frac{\lambda}{2}\), where \(n=0,1,2,3 \ldots \ldots\)
Step-by-step Solution
Detailed explanation
The equation of the standing wave formed due to superposition of \(Y=A \sin (\omega t-k x)\) and \(Y=A \sin (\omega t+k x)\) is given by,
\(Y=[2 A \cos (k x)] \sin (\omega t)\)
Location of the nodes is dictated by condition:
\([2 A \cos (k x)]=0\)
Therefore,
\(\begin{aligned} & k x=\left(\frac{2 \pi}{\lambda}\right) x=(2 n+1) \frac{\pi}{2} \\ & \Rightarrow x=\left(n+\frac{1}{2}\right) \frac{\lambda}{2} ; n=0,1,2, \ldots .\end{aligned}\)
\(Y=[2 A \cos (k x)] \sin (\omega t)\)
Location of the nodes is dictated by condition:
\([2 A \cos (k x)]=0\)
Therefore,
\(\begin{aligned} & k x=\left(\frac{2 \pi}{\lambda}\right) x=(2 n+1) \frac{\pi}{2} \\ & \Rightarrow x=\left(n+\frac{1}{2}\right) \frac{\lambda}{2} ; n=0,1,2, \ldots .\end{aligned}\)
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