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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
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Verified Answer
The correct answer is:
$4 \mathrm{~cm}$
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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