A progressive wave of frequency $500 \mathrm{~Hz}$ is travelling with a velocity of $360 \mathrm{~ms}^{-1}$.…
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 .............
1.2 m
12 m
0.12 m
0.012 m
Solution
$
\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}
$