Question
The speed of wave is 360 m/sec. And the frequency is 500 Hz. and the phase difference between adjacent particle is $60^{\circ} $. What will be the path difference between them?
✓
Answer
$\begin{array}{l}\text {Given :}\quad v=360 m / s \\n=500 Hz \\\phi=60^{\circ}=\pi / 3 \text { Radian }\end{array}$
Let the path difference between adjacent particle $=x$
$\begin{aligned} \text { Phase difference } & =\frac{2 \pi}{\lambda} \times \text { Path difference } \\
\text { Path difference } & =\frac{\lambda}{2 \pi} \times \text { Phase difference } \\ & =\frac{ v / n}{2 \pi} \times \frac{\pi}{3} \\ & =\frac{360}{500} \times \frac{1}{6}=0.12 \text { meter }\end{aligned}$
Let the path difference between adjacent particle $=x$
$\begin{aligned} \text { Phase difference } & =\frac{2 \pi}{\lambda} \times \text { Path difference } \\
\text { Path difference } & =\frac{\lambda}{2 \pi} \times \text { Phase difference } \\ & =\frac{ v / n}{2 \pi} \times \frac{\pi}{3} \\ & =\frac{360}{500} \times \frac{1}{6}=0.12 \text { meter }\end{aligned}$
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