The phase difference between two points is π /3 . If the frequency of waves is 50 Hz , then, the distance between two points is (Given v=330 m s- 1 )

Question

The phase difference between two points is π /3 . If the frequency of waves is 50 Hz , then, the distance between two points is (Given v=330 m s- 1 ) (A) 2.2 m (B) 1.1 m (C) 0.6 m (D) 1.7 m

Tardigrade Admin · Accepted Answer

From the relation, Δ φ=(2 π /λ )× Δ x Longrightarrow Δ x=(λ /2 π )× Δ φ ldots (i) Also, λ =(v/n) ldots (i i) Now, from Eqs.(i) and (ii), we get Δ x=(v/2 π n)× Δ φ Longrightarrow Δ x=(330/2 π × 50)× (π /3) or Δ x=1.1 m

Q. The phase difference between two points is $π /3$ . If the frequency of waves is $50 Hz$ , then, the distance between two points is (Given $v = 330 m s^{- 1}$ )

$2.2 m$

$1.1 m$

$0.6 m$

$1.7 m$

From the relation,
$Δ ϕ = \frac{2 π}{λ} \times Δ x$
$⟹ Δ x = \frac{λ}{2 π} \times Δ ϕ \dots (i)$
Also, $λ = \frac{v}{n} \dots (ii)$
Now, from Eqs.(i) and (ii), we get
$Δ x = \frac{v}{2 πn} \times Δ ϕ$
$⟹ Δ x = \frac{330}{2 π \times 50} \times \frac{π}{3}$
$or Δ x = 1.1 m$

Q. The phase difference between two points is π/3 . If the frequency of waves is 50Hz , then, the distance between two points is (Given v=330ms−1 )

2.2m

1.1m

0.6m

1.7m