A) 1.07rad
B) 2.07 rad
C) 0.5 rad
D) 1.5 rad
Correct Answer: A
Solution :
The given waves are \[{{y}_{1}}={{10}^{-6}}\sin [100t+(x/50)+0.5]m\] and \[{{y}_{2}}={{10}^{-6}}\cos [100t+(x/50)]m\] \[\Rightarrow \]\[{{y}_{2}}={{10}^{-6}}\sin [100t+(x/50)+\frac{\pi }{2}]m\] \[\left[ \because \sin \left( \frac{\pi }{2}+\theta \right)=\cos \theta \right]\] Hence, the phase difference between the waves is \[\Delta \phi =\left( \frac{\pi }{2}-0.5 \right)rad\] \[=\left( \frac{3.14}{2}-0.5 \right)rad\] \[=(1.57-0.5)\text{ }rad\] \[=(1.07)rad\] NOTE: The given waves are sine and cosine. function, so they are plane progressive harmonic waves.
You need to login to perform this action.
You will be redirected in
3 sec
