Waves - Result Question 11
11. The phase difference between two waves, represented by
$y_1=10^{-6} \sin {100 t+(x / 50)+0.5} m$ $y_2=10^{-6} \cos {100 t+(x / 50)} m$ where $x$ is expressed in metres and $t$ is expressed in seconds, is approximately [2004]
(a) 1.5 radians
(b) 1.07 radians
(c) 2.07 radians
(d) 0.5 radians
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Answer:
Correct Answer: 11. (b)
Solution:
- (b) $y_1=10^{-6} \sin (100 t+x / 50+0.5) m$
$=10^{-6} \cos (100 t+x / 50-\pi / 2+$
$0.5) m$
$ \begin{aligned} & y_2=10^{-6} \cos (100 t+x / 50) m \\ & \therefore \quad \phi=\pi / 2-0.5=1.07 rad \end{aligned} $
