A) 1.07 rad
B) 2.07 rad
C) 0.5 rad
D) 1.5 rad
Correct Answer: A
Solution :
The given waves are \[{{y}_{1}}={{10}^{-6}}\sin \left[ 100t+(x/50)+0.5 \right]m\] and \[{{y}_{2}}={{10}^{-6}}\cos \left[ 100t+(x/50) \right]m\] \[\Rightarrow \]\[{{y}_{2}}={{10}^{-6}}\sin \left[ 100t+(x/50)+\frac{\pi }{2} \right]m\] \[\left[ \because \sin \left( \frac{\pi }{2}+\theta \right)=\cos \theta \right]\] Hence, the phase difference between the waves is \[\Delta \phi =\left( \frac{\pi }{2}-0.5 \right)rad=\left( \frac{3.14}{2}-0.5 \right)rad\] \[=(1.57-0.5)rad=(1.07)rad\]
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