Q165 / 170 · Manipal (MET) 2019 · Physics

$y_1=4 \sin (\omega t+k x), \quad y_2=-4 \cos (\omega t+k x)$ the phase difference is

  • A
    $\frac{\pi}{2}$
  • Correct answer
    B
    $\frac{3 \pi}{2}$
  • C
    $\pi$
  • D
    zero

Solution

$y_1=4 \sin (\omega t+k x)$ ...(i) $\begin{aligned} y_2 & =-4 \cos (\omega t+k x) \\ & =-4 \sin (\omega t+k x+\pi / 2)\end{aligned}$ $=4 \sin \left(\omega t+k x+\frac{3 \pi}{2}\right)$ From Eqs. (i) and (ii), we get The phase difference, $\Delta \phi=\frac{3 \pi}{2}$

Hence, the correct answer is option: 'B'.

Relevant Questions