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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


A
1.0 radian
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B
1.25 radian
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C
1.57 radian
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D
0.57 radian
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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, $$\triangle \phi= \phi_2 - \phi_1$$
$$=(\omega t + kx + \dfrac{\pi}{2})- (\omega t + kx + 0.57)$$
$$= \dfrac{\pi}{2} - 0.57 = 1.57 - 0.57= 1$$ radian

Physics

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