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° is ...........

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° is ........... (A) 1.2 m (B) 12 m (C) 0.12 m (D) 0.012 m

Tardigrade Admin · Accepted Answer

f=500 Hz , v=360 ms -1 λ=(v/f)=(360/500) Now, a phase difference of 60° corresponds to a path difference of (60/360) × λ So, distance between 2 particles is d=(60/360) × λ=(60/360) × (360/500)=0.12 m

Q. A progressive wave of frequency $500 Hz$ is travelling with a velocity of $360 m s^{- 1}$ . The distance between the two points, having a phase difference of $6 0^{\circ}$ is ...........

1.2 m

12 m

0.12 m

0.012 m

$f = 500 Hz, v = 360 m s^{- 1}$
$λ = \frac{v}{f} = \frac{360}{500}$
Now, a phase difference of $6 0^{\circ}$ corresponds to
a path difference of $\frac{60}{360} \times λ$
So, distance between 2 particles is
$d = \frac{60}{360} \times λ = \frac{60}{360} \times \frac{360}{500} = 0.12 m$

Q. A progressive wave of frequency 500Hz is travelling with a velocity of 360ms−1. The distance between the two points, having a phase difference of 60∘ is ...........