### Waves - Result Question 2

#### 2. Two waves are represented by the equations $y_1=a \sin (\omega t+kx+0.57) m$ and $y_2=a \cos (\omega t$ $+kx) m$, where $x$ is in meter and $t$ in sec. The phase difference between them is [2011]

(a) 1.0 radian

(b) 1.25 radian

(c) 1.57 radian

(d) 0.57 radian

## Show Answer

#### Answer:

Correct Answer: 2. (a)

Solution:

- (a) Here, $y_1=a \sin (\omega t+k x+0.57)$

so, $\phi_1=0.57$

and $y_2=a \cos (\omega t+k x)$

$ =a \sin [\frac{\pi}{2}+(\omega t+k x)] $

so, $\phi_2=\pi / 2$

Phase difference, $\Delta \phi=\phi_2-\phi_1$

$ \begin{gathered} =\frac{\pi}{2}-0.57=\frac{3.14}{2}-0.57=1.57-0.57 \\ =1 \text{ radian } \end{gathered} $