Question

Two waves are represented by the equations

y1 = a sin (wt + kx + 0.57) m and y2 = a cos (wt + kx)m,

where x is in metres and t is in seconds. The phase difference between them is

• A. 1.0 radian
• B. 1.25 radian
• C. 1.57 radian
• D. 0.57 radian

Answer 1

Answer: (A)

1.0 radian

Answer 2

$y_{1} = a\sin(\omega t + kx + 0.57)$

$\therefore$Phase, $\varphi_{1} = (\omega t + kx + 0.57)$

$y_{2} = a\cos(\omega t + kx) = a\sin(\omega t + kx + \frac{\pi}{2})$

$\therefore$Phase, $\varphi_{2} = \omega t + kx + \frac{\pi}{2}$

Phase difference, $\Delta\varphi = \varphi_{2} - \varphi_{1}$

$= (\omega t + kx + \frac{\pi}{2}) - (\omega t + kx + 0.57)$

$= \frac{\pi}{2} - 0.57 = 1.57 - 0.57 = 1radian$