| The phase difference between two waves, represented by [AIPMT (S) 2004] |
| \[{{y}_{1}}={{10}^{-6}}\sin [100t+(x/50)+0.5]\,m\] |
| \[{{y}_{2}}={{10}^{-6}}\cos [100t+(x/50)]\,m\] |
| where x is expressed in metres and t is expressed in seconds, is approximately: |
A) 1.07 rad
B) 2.07 rad
C) 0.5 rad
D) 1.5 rad
Correct Answer: A
Solution :
| The given waves are |
| \[{{y}_{1}}={{10}^{-6}}\sin [100t+(x/50)+0.5]m\] |
| 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 |
| \[\Delta \phi =\left( \frac{\pi }{2}-0.5 \right)rad\] |
| \[=\left( \frac{3.14}{2}-0.5 \right)\,\,rad\] |
| \[=(1.57-0.5)\,\,rad\] |
| \[=(1.07)\,\,rad\] |
| Note: The given waves are sine and cosine function, so they are plane progressive harmonic waves. |
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