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A progressive wave is given by, $Y=12 \sin (5 t-4 x)$. On this wave, how far away are the two points having a phase difference of $90^{\circ}$ ?
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Verified Answer
The correct answer is:
$\frac{\pi}{8}$
Given, $y=12 \sin (5 t-4 x)$ $\therefore \quad y=12 \sin 2 \pi\left(\frac{5 t}{2 \pi}-\frac{4 x}{2 \pi}\right)$
Comparing above eq. with,
$\mathrm{y}=\mathrm{A} \sin 2 \pi\left(\frac{\mathrm{x}}{\lambda}-\frac{\mathrm{t}}{\mathrm{T}}\right)$
We get, $\lambda=\frac{2 \pi}{4}$
Relation between phase difference and path difference is
$\begin{aligned} \Delta \phi & =\frac{2 \pi}{\lambda} \Delta \mathrm{x} \\ \therefore \quad \frac{\pi}{2} & =\frac{2 \pi}{\left(\frac{2 \pi}{4}\right)} \Delta \mathrm{x} \\ \therefore \quad \Delta \mathrm{x} & =\frac{\pi}{8}\end{aligned}$
Comparing above eq. with,
$\mathrm{y}=\mathrm{A} \sin 2 \pi\left(\frac{\mathrm{x}}{\lambda}-\frac{\mathrm{t}}{\mathrm{T}}\right)$
We get, $\lambda=\frac{2 \pi}{4}$
Relation between phase difference and path difference is
$\begin{aligned} \Delta \phi & =\frac{2 \pi}{\lambda} \Delta \mathrm{x} \\ \therefore \quad \frac{\pi}{2} & =\frac{2 \pi}{\left(\frac{2 \pi}{4}\right)} \Delta \mathrm{x} \\ \therefore \quad \Delta \mathrm{x} & =\frac{\pi}{8}\end{aligned}$
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