Question

# Two waves are represented by the equations $$y_1= asin(\omega t + kx + 0.57) m$$ and$$y_2= acos(\omega t + kx) m$$, Where $$x$$ is in meters and $$t$$ is in seconds. The phase difference between them is then

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Solution

## The correct option is A $$1.0$$ radian$$y_1= asin (\omega t + kx + 0.57)$$$$\therefore Phase, \phi_1 = (\omega t + kx + \dfrac{\pi}{2})$$$$y_2=acos(\omega t + kx)= asin (\omega t + kx + \dfrac{\pi}{2})$$$$\therefore Phase, \phi_2= \omega t + kx + \dfrac{\pi}{2}$$ Phase different, $$\triangle \phi= \phi_2 - \phi_1$$$$=(\omega t + kx + \dfrac{\pi}{2})- (\omega t + kx + 0.57)$$$$= \dfrac{\pi}{2} - 0.57 = 1.57 - 0.57= 1$$ radianPhysics

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