MHT CET · Physics · Waves and Sound
A stationary wave is represented by \(y=10 \sin \left(\frac{\pi x}{4}\right) \cos (20 \pi t)\) where \(x\) and \(y\) are in \(\mathrm{cm}\) and \(t\) in second. The distance between two consecutive nodes is
- A \(1 \mathrm{~cm}\)
- B \(8 \mathrm{~cm}\)
- C \(4 \mathrm{~cm}\)
- D \(2 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(C) \(4 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
Given, \(y=10 \sin \left(\frac{\pi x}{4}\right) \cos (\pi t)\)
Compare with standard equation formed after superposing
\(\begin{aligned} & y=A \sin (k x+\omega t) \text { and } y=A \sin (k x-\omega t) \\ & y=2 A \sin (k x) \cos (\omega t)\end{aligned}\)
Therefore, \(k=\frac{\pi}{4}\)
We know, the propagation constant is:
\(k=\frac{2 \pi}{\lambda}=\frac{\pi}{4}\)
\(\Rightarrow \lambda=8 \mathrm{~cm}\)
The distance between two consecutive nodes is given by half of wavelength:
\(\frac{\lambda}{2}=4 \mathrm{~cm}\)
Compare with standard equation formed after superposing
\(\begin{aligned} & y=A \sin (k x+\omega t) \text { and } y=A \sin (k x-\omega t) \\ & y=2 A \sin (k x) \cos (\omega t)\end{aligned}\)
Therefore, \(k=\frac{\pi}{4}\)
We know, the propagation constant is:
\(k=\frac{2 \pi}{\lambda}=\frac{\pi}{4}\)
\(\Rightarrow \lambda=8 \mathrm{~cm}\)
The distance between two consecutive nodes is given by half of wavelength:
\(\frac{\lambda}{2}=4 \mathrm{~cm}\)
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