The phase difference between two points separated by 0.8 m , in a wave of frequency 120 Hz , is (π /2) . The velocity of wave is

Question

The phase difference between two points separated by 0.8 m , in a wave of frequency 120 Hz , is (π /2) . The velocity of wave is (A)  720 ms-1  (B)  384 ms-1 (C)  250 ms-1 (D)  1 ms-1

Tardigrade Admin · Accepted Answer

It is given that v = 120 Hz, phase difference Δ φ = (π/2) and path difference Δ x = 0.8 m. As Δ φ = (2 π/λ).Δ x ⇒ λ = (2 π/Δ φ) . Δ x = (2 π × 0.8/π / 2) = 3.2 m Hence, wave velocity v = v λ= 120 × 3.2 = 384 ms-1 .

Q. The phase difference between two points separated by $0.8 m$ , in a wave of frequency $120 Hz$ , is $\frac{π}{2}$ . The velocity of wave is

$720 m s^{- 1}$

$384 m s^{- 1}$

$250 m s^{- 1}$

$1 m s^{- 1}$

It is given that $v = 120 Hz,$
phase difference $Δ ϕ = \frac{π}{2}$ and path difference $Δ x = 0.8 m .$
As $Δ ϕ = \frac{2 π}{λ} .Δ x$
$\Rightarrow λ = \frac{2 π}{Δ ϕ} .Δ x = \frac{2 π \times 0.8}{π /2} = 3.2 m$
Hence, wave velocity
$v = v λ = 120 \times 3.2 = 384 m s^{- 1}$ .

Q. The phase difference between two points separated by 0.8m , in a wave of frequency 120Hz , is 2π​ . The velocity of wave is

720ms−1

384ms−1

250ms−1

1ms−1