Question

A progressive wave of frequency $500\, Hz$ is travelling with a velocity of $360\, ms ^{-1}$. The distance between the two points, having a phase difference of $60^{\circ}$ is ...........

• A. 1.2 m
• B. 12 m
• C. 0.12 m
• D. 0.012 m

Answer 1

Answer: (C)

$f=500\, Hz ,\, v=360\, ms ^{-1}$
$\lambda=\frac{v}{f}=\frac{360}{500}$
Now, a phase difference of $60^{\circ}$ corresponds to
a path difference of $\frac{60}{360} \times \lambda$
So, distance between 2 particles is
$d=\frac{60}{360} \times \lambda=\frac{60}{360} \times \frac{360}{500}=0.12\, m$