Two waves are given by y1 = cos (4t - 2x) and y2 = sin big(4t - 2x + (π/4) big). The phase difference between the two waves is

Question

Two waves are given by y1 = cos (4t - 2x) and y2 = sin big(4t - 2x + (π/4) big). The phase difference between the two waves is (A) (π/4) (B) (- π/4) (C) (3 π/4) (D) (π/2)

Tardigrade Admin · Accepted Answer

Equations of waves y1 = cos (4t - 2x) = sin big(4t - 2x + (π/2) big) and y2 = sin big(4t - 2x + (π/4) big) Therefore, phase difference between the two waves is Δ φ = big(4t - 2x + (π/4) big) - big(4t - 2x + (π/2) big) = (π/4) - (π/2) = - (π/4)

Q. Two waves are given by $y_{1} = cos (4 t - 2 x)$ and $y_{2} = s in (4 t - 2 x + \frac{π}{4}) .$ The phase difference between the two waves is

$\frac{π}{4}$

$\frac{- π}{4}$

$\frac{3 π}{4}$

$\frac{π}{2}$

$\frac{3 π}{2}$

Equations of waves
$y_{1} = cos (4 t - 2 x) = sin (4 t - 2 x + \frac{π}{2})$
and $y_{2} = sin (4 t - 2 x + \frac{π}{4})$
Therefore, phase difference between the two waves is
$Δ ϕ = (4 t - 2 x + \frac{π}{4}) - (4 t - 2 x + \frac{π}{2})$
$= \frac{π}{4} - \frac{π}{2} = - \frac{π}{4}$

Q. Two waves are given by y1​=cos(4t−2x) and y2​=sin(4t−2x+4π​). The phase difference between the two waves is

4π​

4−π​

43π​

2π​

23π​

Equations of waves y1​=cos(4t−2x)=sin(4t−2x+2π​) and y2​=sin(4t−2x+4π​) Therefore, phase difference between the two waves is Δϕ=(4t−2x+4π​)−(4t−2x+2π​) =4π​−2π​=−4π​