The path difference between the two waves y1=a1 sin (ω t-(2 π x/λ)) and y2=a2 cos (ω t-(2 π x/λ)+φ) is

Question

The path difference between the two waves y1=a1 sin (ω t-(2 π x/λ)) and y2=a2 cos (ω t-(2 π x/λ)+φ) is (A)  (λ /2π )φ  (B)  (λ /2π )( φ +(π /2) )  (C)  (2π /λ )( φ -(π /2) )  (D)  (2π /λ )φ

Tardigrade Admin · Accepted Answer

y1 =a1 sin (ω t-(2 π x/λ)) and y2 =a2 ⋅ cos (ω t-(2 π x/λ)+φ) =a2 sin (ω t-(2 π x/λ)+φ+(π/2)) So, phase difference =φ+(π/2) and path difference Δ=(λ/2 π)(φ+(π/2))

Q. The path difference between the two waves $y_{1} = a_{1} sin (ω t - \frac{2 π x}{λ})$ and $y_{2} = a_{2} cos (ω t - \frac{2 π x}{λ} + ϕ)$ is

$\frac{λ}{2 π} ϕ$

$\frac{λ}{2 π} (ϕ + \frac{π}{2})$

$\frac{2 π}{λ} (ϕ - \frac{π}{2})$

$\frac{2 π}{λ} ϕ$

$y_{1} = a_{1} sin (ω t - \frac{2 π x}{λ})$ and
$y_{2} = a_{2} \cdot cos (ω t - \frac{2 π x}{λ} + ϕ)$
$= a_{2} sin (ω t - \frac{2 π x}{λ} + ϕ + \frac{π}{2})$
So, phase difference $= ϕ + \frac{π}{2}$
and path difference $Δ = \frac{λ}{2 π} (ϕ + \frac{π}{2})$

Q. The path difference between the two waves y1​=a1​sin(ωt−λ2πx​) and y2​=a2​cos(ωt−λ2πx​+ϕ) is

2πλ​ϕ

2πλ​(ϕ+2π​)

λ2π​(ϕ−2π​)

λ2π​ϕ