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

Question

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

Tardigrade Admin · Accepted Answer

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

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

$\frac{π}{4}$

$- \frac{π}{4}$

$\frac{3 π}{4}$

$\frac{π}{2}$

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​

−4π​

43π​

2π​