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A progressive wave of frequency $500 \mathrm{~Hz}$ is travelling with a velocity of $360 \mathrm{~ms}^{-1}$. The distance between the two points, having a phase difference of $60^{\circ}$ is .............
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The correct answer is:
0.12 m
$$
\begin{gathered}
f=500 \mathrm{~Hz}, v=360 \mathrm{~ms}^{-1} \\
\lambda=\frac{v}{f}=\frac{360}{500}
\end{gathered}
$$
Now, a phase difference of $60^{\circ}$ corresponds to a path difference of $\frac{60}{360} \times \lambda$.
So, distance between 2 particles is
$$
d=\frac{60}{360} \times \lambda=\frac{60}{360} \times \frac{360}{500}=0.12 \mathrm{~m}
$$
\begin{gathered}
f=500 \mathrm{~Hz}, v=360 \mathrm{~ms}^{-1} \\
\lambda=\frac{v}{f}=\frac{360}{500}
\end{gathered}
$$
Now, a phase difference of $60^{\circ}$ corresponds to a path difference of $\frac{60}{360} \times \lambda$.
So, distance between 2 particles is
$$
d=\frac{60}{360} \times \lambda=\frac{60}{360} \times \frac{360}{500}=0.12 \mathrm{~m}
$$
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